Determining Present Value for Future Financial Goals
Word equations to
balanced chemical equations drill
2b
Each of the following slides first gives you a word
equation.
You must write the balanced chemical equation, with
phases, then click ahead to check your work.
Each slide is numbered, so you can do some, come
back and then do more.
Practice makes perfect.
Aluminum and oxygen form aluminum oxide
Aluminum and oxygen form aluminum oxide
_Al
(S)
+ _O
2(G)
→ _Al
2
O
3(S)
SKELETON
Al
+3
cation and O
-2
anion,
criss cross
to get the right formula.
Aluminum and oxygen form aluminum oxide
_Al
(S)
+ _O
2(G)
→ _Al
2
O
3(S)
SAY: one aluminum up front in reactants, but
two aluminums in the product, put a 2 on the
reactant side to start.
Aluminum and oxygen form aluminum oxide
2
Al
(S)
+ _O
2(G)
→ _Al
2
O
3(S)
SAY: two aluminums up front in reactants, and
two aluminums in the product, good.
SAY: there are 2 oxygens in the reactants, but three in
the products. Even 2 and Odd 3, bad.
To fix this, we double the odd one, by doubling up Al
2
O
3
Aluminum and oxygen form aluminum oxide
2Al
(S
)
+ _O
2(G)
→
2
Al
2
O
3(S)
SAY: two aluminums up front in reactants, and
four aluminums in the product, bad.
To fix this we must change the 2Al in the reactants to a four.
Aluminum and oxygen form aluminum oxide
2Al
(S)
+ _O
2(G)
→
2Al
2
O
3(S)
SAY: now we’ve got 2 Al in reactants, but 4 in
products, so change the first “2” into a four, then
start again.
4
Aluminum and oxygen form aluminum oxide
4Al
(S)
+ _O
2(G)
→
2Al
2
O
3(S)
SAY: 4 Al in reactants, and 4 Al in products
(good). Only 2 oxygen in reactants, but SIX in
product. HOW DO WE GET 6 OXGYEN in
reactants?
2Al
2
O
3
Means,
Al
2
O
3
and another
Al
2
O
3
That’s 6
oxygens!
Aluminum and oxygen form aluminum oxide
4Al
(S)
+
3
O
2(G)
→ 2Al
2
O
3(S)
Put a 3 in front of O
2
, and now it’s balanced.
It’s a 4:3:2 ratio, simple whole numbers, just like
our friend John Dalton, and your teacher like.
All good.
Pentane gas (C
5
H
10
) combusts
Pentane combusts
_C
5
H
10(G)
+ _O
2(G)
→ _CO
2(G)
+ _H
2
O
(G)
Combustion is ALWAYS a hydrocarbon –
in this case the pentane – combining
with pure oxygen, and then ALWAYS
forming carbon dioxide and water.
SKELETON
Pentane combusts
_C
5
H
10(G)
+ _O
2(G)
→
5
CO
2(G)
+ _H
2
O
(G)
SAY, 5 carbons in pentane, so put a 5
in front of CO
2
to keep the carbons
balanced.
Pentane combusts
_
C
5
H
10(G)
+ _O
2(G)
→ 5CO
2(G)
+
5
H
2
O
(G)
Next, there are 10 hydrogens in
pentane, we need to put a 5 in front of
water, to make the hydrogens balance.
Pentane combusts
_
C
5
H
10(G)
+ _O
2(G)
→ 5CO
2(G)
+ 5H
2
O
(G)
There are always 2
oxygens to start a
combustion, how many
on the product side?
Ten in CO
2
and then 5 in water.
10 + 5 = 15 (that’s ODD and bad)
The odd is in the water, so our only
hope is to DOUBLE the waters, and
start over.
Change the 5 to a 10.
Pentane combusts
_
C
5
H
10(G)
+ _O
2(G)
→ 5CO
2(G)
+
10
H
2
O
(G)
There are always 2
oxygens to start a
combustion, how many
on the product side?
Ten in CO
2
and then 5 in water.
10 + 5 = 15 (that’s ODD and bad)
The odd is in the water, so our only
hope is to DOUBLE the waters, and
start over.
Change the 5 to a 10.
Pentane combusts
2
C
5
H
10(G)
+ _O
2(G)
→ 5CO
2(G)
+ 10H
2
O
(G)
Still 5 carbon in pentane and 5 CO
2
(good so far).
10 H in pentane, but now 20 H in the water. Put a 2
in front of pentane to balance the “H”, start again.
Pentane combusts
2C
5
H
10(G)
+ _O
2(G)
→
10
CO
2(G)
+ 10H
2
O
(G)
SAY: 10 C in pentane, only 5 in CO
2
, fix that five into
a 10, then start over again. (yeohhh! Crazy huh?)
Pentane combusts
2C
5
H
10(G)
+ _O
2(G)
→ 10CO
2(G)
+ 10H
2
O
(G)
SAY: 10 C in pentane, 10 in CO
2
, good.
20 H in the pentanes, and 20 H in the water, good!
Only 2 oxygens in pure oxygen, but 20 in CO
2
and 10
more in the water. (20 + 10 = 30). How do we get 30
in the pure oxygen?
Pentane combusts
2C
5
H
10(G)
+ 15O
2(G)
→ 10CO
2(G)
+ 10H
2
O
(G)
Change the pure oxygen to a
15O
2
(15 x 2 = 30) now balanced.
Rubidium metal is put into calcium nitrate solution
Rubidium metal is put into calcium nitrate solution
_Rb
(S)
+ _Ca(NO
3
)
2 (AQ)
→ _RbNO
3(AQ)
+ _Ca
(S)
SKELETON
Calcium makes a Ca
+2
cation
Nitrate is on table E, NO
3
-1
anion
Criss cross to get proper formula
Rubidium makes a Rb
+1
cation
Nitrate is on table E, NO
3
-1
anion
Criss cross to get proper formula
Rubidium metal is put into calcium nitrate solution
_Rb
(S)
+ _Ca(NO
3
)
2 (AQ)
→
2
RbNO
3(AQ)
+_ Ca
(S)
SAY: 1 Rb in reactants, and 1 Rb in products, so far so good.
Then… 1 Ca in reactants, and 1 a in products, also good.
But… two nitrates in the reactant side, so DOUBLE the
RbNO
3
to double nitrates in the products (then start over).
Rubidium metal is put into calcium nitrate solution
2
Rb
(S)
+ _Ca(NO
3
)
2 (AQ)
→
2
RbNO
3(AQ)
+_ Ca
(S)
SAY: 1 Rb in reactants, and 2 Rb in products, put a 2 in front
of Rb in the reactants.
Rubidium metal is put into calcium nitrate solution
2
Rb
(S)
+ _Ca(NO
3
)
2 (AQ)
→
2
RbNO
3(AQ)
+_ Ca
(S)
SAY: Rb good 2:2
Ca good at 1:1
nitrates good at 2:2
Balanced! In a 2:1:2:1 ratio –
John Dalton would be happy, I’m happy.
You be happy too.
Rubidium metal is put into calcium nitrate solution
2Rb
(S)
+ Ca(NO
3
)
2 (AQ)
→ 2RbNO
3(AQ)
+ Ca
(S)
Why you might have got this one wrong:
Rubidium makes only a +1 cation so forming rubidium nitrate
can only be in a 1:1 ratio. The “2” in the calcium nitrate needs
to be BALANCED away, not by you sticking a 2 in as in:
Rb(NO
3
)
2
because it’s easier to do.
This: Rb(NO
3
)
2
is just not possible.
Cobalt (III) chlorate solution and magnesium
chromate solutions are mixed together
Cobalt (III) chlorate solution and magnesium
chromate solutions are mixed together
_Co(ClO
3
)
3 (AQ)
+ _MgCrO
4(AQ)
→
SKELETON
This is the set up for a double replacement reaction, we’re
starting with 2 AQ solutions.
Don’t just do this in your head, in our class we will make the switch, and balance the
equation. If by chance both products are AQ also, that means it was NO REACTION.
For practice balancing, we always go through the motions. You can’t decide no
reaction until it’s balanced and THEN you go to table F to find the solid precipitate, if
it’s even there.
Cobalt (III) chlorate solution and magnesium
chromate solutions are mixed together
_Co(ClO
3
)
3 (AQ)
+ _MgCrO
4(AQ)
→ _Co
CrO
4(?)
+ _MgClO
3(?)
Switch the anions around, here the chromate anion starts with
the Mg, but ends up with the Co. The chlorate anion starts
with Co, but ends up with the Mg. It’s a double switch (like a
square dance, switch your partners, do si do).
FIX IN NEXT STEP!
Note, we have NO IDEA YET about the solubility of the products, nor would we look before
balancing the equation. Switch, fix, balance, and then table “F” them.
Cobalt (III) chlorate solution and magnesium
chromate solutions are mixed together
_Co(ClO
3
)
3 (AQ)
+ _MgCrO
4(AQ)
→ _
Co
CrO
4
(?)
+ _
MgClO
3
(?)
We switched ‘em.
Now… FIX ‘EM!
Co
+3
and CrO
4
-2
gets criss-crossed to : Co
2
(CrO
4
)
3
Mg
+2
and ClO
3
-1
gets criss-crossed to: Mg(ClO
3
)
2
_Co(ClO
3
)
3 (AQ)
+ _MgCrO
4(AQ)
→
Co
2
(CrO
4
)
3(?)
+
Mg(ClO
3
)
2(?)
Cobalt (III) chlorate solution and magnesium
chromate solutions are mixed together
2Co(ClO
3
)
3 (AQ)
+ _MgCrO
4(AQ)
→ _
Co
2
(CrO
4
)
3(?)
+
_
Mg(ClO
3
)
2(?)
We switched ‘em, we fixed ‘em… now balance ‘em!
SAY: one Co in reactants, but two in products…
Put a 2 in front of Co(ClO
3
)
3
Cobalt (III) chlorate solution and magnesium
chromate solutions are mixed together
2Co(ClO
3
)
3 (AQ)
+ _MgCrO
4(AQ)
→ _Co
2
(CrO
4
)
3(?)
+
3Mg(ClO
3
)
2(?)
SAY: 2 Co in reactants, and 2 in products… good!
SIX chlorates, in reactants, two in products,
Put a 3 in front of Mg(ClO
3
)
2
to get six chlorates in products, start over.
Cobalt (III) chlorate solution and magnesium
chromate solutions are mixed together
2Co(ClO
3
)
3 (AQ)
+
3
MgCrO
4(AQ)
→ _Co
2
(CrO
4
)
3(?)
+ 3Mg(ClO
3
)
2(?)
SAY: 2 Co in reactants, and 2 in products… good!
SIX chlorates, in reactants, SIX in products… .good!
Only 1 Mg in reactants, but 3 are in the products,
so put a 3 in front of MgCrO
4
, start over again.
Cobalt (III) chlorate solution and magnesium
chromate solutions are mixed together
2Co(ClO
3
)
3 (AQ)
+ 3MgCrO
4(AQ)
→ _Co
2
(CrO
4
)
3(?)
+ 3Mg(ClO
3
)
2(?)
SAY: 2 Co in reactants, and 2 in products… good!
SIX chlorates, in reactants, SIX in products… .good!
3 Mg in reactants, and 3 are in the products… good!
3 chromates in reactants, and LOOK, 3 chromates in products, balanced!
This is a 2:3:1:3 ratio, it can’t be reduced so - John Dalton is happy!
Cobalt (III) chlorate solution and magnesium
chromate solutions are mixed together
2Co(ClO
3
)
3 (AQ)
+ 3MgCrO
4(AQ)
→ _Co
2
(CrO
4
)
3(
S
)
+ 3Mg(ClO
3
)
2(
AQ
)
LAST STEP: check products on table F, which is AQ or S???
If your last name is chromate (right side) you are insoluble but there are
exceptions – like Mg, but not Co, so the first product is SOLID! A
precipitate is the proof that a DR happened!
If your last name is chlorate, you are soluable (AQ) without exception!
DONE finally!
Hydrogen mono-iodide gas decomposes
Hydrogen monoiodide gas decomposes
_HI
(G)
→ _H
2(G)
+ _I
2(S)
Hydrogen mono-ioidide is a 1:1 ratio, and the both products
are HONClBrIF twins, diatomic.
The reactant is a gas, we’re told that. Hydrogen is a gas
normally, and iodine is a solid at normal temperatures.
SKELETON
Hydrogen monoiodide gas decomposes
2
HI
(G)
→ _H
2(G)
+ _I
2(S)
SAY: One H in reactant, two in products.
Put a 2 in front of HI, and start over.
SKELETON
Hydrogen monoiodide gas decomposes
2
HI
(G)
→ _H
2(G)
+ _I
2(S)
SAY: 2 H’s in reactants, and two in products, good.
SAY: 2 I’s in reactants, and also in products.
Balanced! (this one was easy)
Carbon and hydrogen form ethane (C
2
H
6
) gas
Carbon and hydrogen form ethane (C
2
H
6
) gas
_C
(S)
+ _H
2(G)
→ _C
2
H
6(G)
Carbon is a solid and it is atomic, just “C”
Hydrogen is a gas, and is diatomic HONClBrIF Twin
You’re given the formula for ethane and it’s a gas.
SKELETON
Carbon and hydrogen form ethane (C
2
H
6
) gas
2
C
(S)
+ _H
2(G)
→ _C
2
H
6(G)
SAY: one C in reactant, but two in products, put a 2 in
front of carbon, start over.
Carbon and hydrogen form ethane (C
2
H
6
) gas
2
C
(S)
+
3
H
2(G)
→ _C
2
H
6(G)
SAY: two C in reactants, and two in products, good.
SAY: 2 H’s in reactants, but 6 H’s in products. How
to get six H’s in front? Add a three in front of the pure
hydrogen. Start over again!
Carbon and hydrogen form ethane (C
2
H
6
) gas
2
C
(S)
+
3
H
2(G)
→ _C
2
H
6(G)
SAY: two C in reactants, and two in products, good.
SAY: 6 H’s in reactants, but 6 H’s in products. Good!
Balanced, this was easier also. In a 2:3:1 ratio (JD
says yay!
Propanol combusts (C
3
H
7
OH)
Propanol is a type of alcohol, it’s an oxygenated hydrocarbon.
That means it is made up of hydrogen and carbon (like a
hydrocarbon) PLUS some oxygen. This is fine, but it makes
balancing a bit trickier, as now there is oxygen in both
reactants as well as both products, but you’re smart, this is
fun.
Propanol combusts (C
3
H
7
OH)
_C
3
H
7
OH
(G)
+ _O
2(G)
→ _CO
2(G)
+ _H
2
O
(G)
In combustion, a hydrocarbon, or OXYGENTATED HYDROCARBON
combines rapidly with oxygen and forms only carbon dioxide and water.
Pure oxygen is diatomic, and the product formulas are well known
already.
SKELETON
Propanol combusts (C
3
H
7
OH)
_C
3
H
7
OH
(G)
+ _O
2(G)
→
3
CO
2(G)
+ _H
2
O
(G)
SAY: 3 carbons in the reactants, so put a 3 in front of CO
2
, start over.
Propanol combusts (C
3
H
7
OH)
_C
3
H
7
OH
(G)
+ _O
2(G)
→
3
CO
2(G)
+
_
H
2
O
(G)
SAY: 3 carbons in the reactants, and 3 in products, good.
Propanol combusts (C
3
H
7
OH)
_C
3
H
7
OH
(G)
+ _O
2(G)
→
3
CO
2(G)
+
4
H
2
O
(G)
SAY: 3 carbons in the reactants, and 3 in products, good.
SAY: 7 + 1 H = 8H in reactants, we need to put a 4 in front of the water.
Start over.
Propanol combusts (C
3
H
7
OH)
_C
3
H
7
OH
(G)
+ _O
2(G)
→
3
CO
2(G)
+
4
H
2
O
(G)
SAY: 3 carbons in the reactants, and 3 in products, good.
SAY: 7 + 1 H = 8H in reactants, and 8 H in products, good.
SAY: two oxygens in reactants, but 6 + 4 = 10 oxygen in products
(NOT GOOD).
To fix this, think… we need even oxygen on reactant side, we need to
double the propanol to do that, then start over.
Propanol combusts (C
3
H
7
OH)
2
C
3
H
7
OH
(G)
+ _O
2(G)
→
3
CO
2(G)
+
4
H
2
O
(G)
SAY: 6 carbons in the reactants, and 3 in products, to fix this,
change the 3 into a 6, and start over.
6
Propanol combusts (C
3
H
7
OH)
2
C
3
H
7
OH
(G)
+ _O
2(G)
→
6
CO
2(G)
+
4
H
2
O
(G)
SAY: 6 carbons in the reactants, and 3 in products, good.
SAY 2 x (7+1) H’s in reactants = 16 H, and only 8 in products, to fix this
change the four into an 8 and start over.
8
Propanol combusts (C
3
H
7
OH)
2
C
3
H
7
OH
(G)
+ _O
2(G)
→
6
CO
2(G)
+
8
H
2
O
(G)
SAY: 6 carbons in the reactants, and 3 in products, good.
SAY 2 x (7+1) H’s in reactants = 16 H, and now 16 H’s in products too.
We have 2+2 oxygens in reactants, but 12 + 8 = 20 in products.
To fix this we do not want to change propanol again, it wrecks
everything. How can we get 20 oxygens but keep just 2 propanol?
Make the pure oxygen 18 oxygens, but adding a 9!
Propanol combusts (C
3
H
7
OH)
2
C
3
H
7
OH
(G)
+
9
O
2(G)
→
6
CO
2(G)
+
8
H
2
O
(G)
SAY: 6 carbons in the reactants, and 3 in products, good.
SAY 2 x (7+1) H’s in reactants = 16 H, and now 16 H’s in products too.
SAY: we have 2 + 18 Oxygens = 20 in reactants, and we have
12 + 8 oxygens = 20 oxygens in the products,
balanced!
Bromine is poured into lithium iodide solution
Bromine is poured into lithium iodide solution
_Br
2(L)
+ _LiI
(AQ)
→
_LiBr
(AQ)
+ _I
2(S)
Bromine is diatomic HONClBrIF twin.
Li makes the Li
+1
cation. Iodine makes only the I
-1
anion, 1:1 ratio.
Li makes the Li
+1
cation. Bromine makes only the Br
-1
anion, 1:1 ratio.
Iodine is diatomic HONClBrIF twin.
SKELETON
Bromine is poured into lithium iodide solution
_Br
2(L)
+ _LiI
(AQ)
→
2
LiBr
(AQ)
+ _I
2(S)
SAY: 2 bromines in reactants, so put a 2 in front of LiBr, start over.
Bromine is poured into lithium iodide solution
_Br
2(L)
+
2
LiI
(AQ)
→ 2LiBr
(AQ)
+ _I
2(S)
SAY: 2 bromines in reactants, 2 in products, good.
SAY: 1 lithium in reactant, but two in products,
put a 2 in front of the LiI and start over again.
Bromine is poured into lithium iodide solution
_Br
2(L)
+ 2LiI
(AQ)
→ 2LiBr
(AQ)
+ _I
2(S)
SAY: 2 bromines in reactants, 2 in products, good.
SAY: 2 lithiums in reactant, but two in products, good.
SAY: 2 iodines in reactants, and we have 2 already in products (great!)
This is already balanced! It’s a 1:2:2:1 ratio, me and JD say yay.
Barium nitrate solution is mixed with
sodium sulfate solution
Here we have two AQ solutions to start, the perfect set up for a
double replacement reaction.
We will write the reactants, then we will write the products.
Barium nitrate solution is mixed with
sodium sulfate solution
_
Ba(NO
3
)
2(AQ
)
+
_Na
2
SO
4(AQ)
→
SKELETON
Ba
+2
and NO
3
-1
get criss-crossed to… Ba(NO
3
)
2
Na
+1
and SO
4
-2
gets criss-crossed to… Na
2
SO
4
Both are aqueous because they are solutions, but you
could check table F to be sure…
If your last name is
nitrate, you’re always aqueous.
If your first name is
sodium, you are always aqueous as well.
Barium nitrate solution is mixed with
sodium sulfate solution
_Ba(NO
3
)
2(AQ)
+ _Na
2
SO
4(AQ)
→
_NaNO
3(?)
+ _BaSO
4(?)
Switch up the anions, to get barium with sulfate, and
sodium with the nitrate. Just switch them now, we’ll
fix them in a minute.
As it turns out,
the NaNO
3
is in a 1:1 ratio +1 cation with the -1 anion.
The BaSO
4
is also a 1:1 ratio, a +2 cation and a -2 anion.
Barium nitrate solution is mixed with
sodium sulfate solution
_Ba(NO
3
)
2(AQ)
+ _Na
2
SO
4(AQ)
→
2
NaNO
3(?)
+ _BaSO
4(?)
SAY: one Ba in reactants, and one in products, good.
SAY: two nitrates in reactants, put a 2 in front of the
sodium nitrate to balance the nitrates, start over.
Barium nitrate solution is mixed with
sodium sulfate solution
_Ba(NO
3
)
2(AQ)
+ _Na
2
SO
4(AQ)
→
2
NaNO
3(?)
+ _BaSO
4(?)
SAY: one Ba in reactants, and one in products, good.
SAY: 2 nitrates in reactants, and 2 in products, good.
SAY: 2 sodiums in reactants, and 2 in products, good.
SAY: one sulfate in reactants and 1 in products, yay!
Balanced with a 1:1:2:1 ratio. Me and J.D. are okay now.
Barium nitrate solution is mixed with
sodium sulfate solution
_Ba(NO
3
)
2(AQ)
+ _Na
2
SO
4(AQ)
→
2NaNO
3(AQ)
+
_BaSO
4(S)
Both reactants are AQ, check the products…
If your first name is sodium, you are AQ.
If your last name is nitrate, you are AQ too.
If your last name is sulfate, you are AQ (with exceptions, and
Ba+2 is an exception, so this product is S, solid! (precipitate,
and proof that a DR happened).
DONE
Learn how to calculate the present value of investments to achieve future financial goals. Understand the concepts with examples and formulas for both single deposit and periodic deposit investments. Practice assignments to strengthen your understanding.
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Word equations to balanced chemical equations drill 2b Each of the following slides first gives you a word equation. You must write the balanced chemical equation, with phases, then click ahead to check your work. Each slide is numbered, so you can do some, come back and then do more. Practice makes perfect.
SKELETON Aluminum and oxygen form aluminum oxide _Al(S) + _O2(G) _Al2O3(S) Al+3 cation and O-2 anion, criss cross to get the right formula.
Aluminum and oxygen form aluminum oxide _Al(S) + _O2(G) _Al2O3(S) SAY: one aluminum up front in reactants, but two aluminums in the product, put a 2 on the reactant side to start.
Aluminum and oxygen form aluminum oxide 2Al(S) + _O2(G) _Al2O3(S) SAY: two aluminums up front in reactants, and two aluminums in the product, good. SAY: there are 2 oxygens in the reactants, but three in the products. Even 2 and Odd 3, bad. To fix this, we double the odd one, by doubling up Al2O3
Aluminum and oxygen form aluminum oxide 2Al(S) + _O2(G) 2Al2O3(S) SAY: two aluminums up front in reactants, and four aluminums in the product, bad. To fix this we must change the 2Al in the reactants to a four.
Aluminum and oxygen form aluminum oxide 2Al(S) + _O2(G) 2Al2O3(S) SAY: now we ve got 2 Al in reactants, but 4 in products, so change the first 2 into a four, then start again.
Aluminum and oxygen form aluminum oxide 4Al(S) + _O2(G) 2Al2O3(S) 2Al2O3 Means,Al2O3 and another Al2O3 That s 6 oxygens! SAY: 4 Al in reactants, and 4 Al in products (good). Only 2 oxygen in reactants, but SIX in product. HOW DO WE GET 6 OXGYEN in reactants?
Aluminum and oxygen form aluminum oxide 4Al(S) + 3O2(G) 2Al2O3(S) Put a 3 in front of O2, and now it s balanced. It s a 4:3:2 ratio, simple whole numbers, just like our friend John Dalton, and your teacher like. All good.
SKELETON Pentane combusts _C5H10(G) + _O2(G) _CO2(G) + _H2O(G) Combustion is ALWAYS a hydrocarbon in this case the pentane combining with pure oxygen, and then ALWAYS forming carbon dioxide and water.
Pentane combusts _C5H10(G) + _O2(G) 5CO2(G) + _H2O(G) SAY, 5 carbons in pentane, so put a 5 in front of CO2 to keep the carbons balanced.
Pentane combusts _C5H10(G) + _O2(G) 5CO2(G) + 5H2O(G) Next, there are 10 hydrogens in pentane, we need to put a 5 in front of water, to make the hydrogens balance.
Pentane combusts _C5H10(G) + _O2(G) 5CO2(G) + 5H2O(G) Ten in CO2 and then 5 in water. There are always 2 oxygens to start a combustion, how many on the product side? 10 + 5 = 15 (that s ODD and bad) The odd is in the water, so our only hope is to DOUBLE the waters, and start over. Change the 5 to a 10.
Pentane combusts _C5H10(G) + _O2(G) 5CO2(G) + 10H2O(G) Ten in CO2 and then 5 in water. There are always 2 oxygens to start a combustion, how many on the product side? 10 + 5 = 15 (that s ODD and bad) The odd is in the water, so our only hope is to DOUBLE the waters, and start over. Change the 5 to a 10.
Pentane combusts 2C5H10(G) + _O2(G) 5CO2(G) + 10H2O(G) Still 5 carbon in pentane and 5 CO2 (good so far). 10 H in pentane, but now 20 H in the water. Put a 2 in front of pentane to balance the H , start again.
Pentane combusts 2C5H10(G) + _O2(G) 10CO2(G) + 10H2O(G) SAY: 10 C in pentane, only 5 in CO2, fix that five into a 10, then start over again. (yeohhh! Crazy huh?)
Pentane combusts 2C5H10(G) + _O2(G) 10CO2(G) + 10H2O(G) SAY: 10 C in pentane, 10 in CO2, good. 20 H in the pentanes, and 20 H in the water, good! Only 2 oxygens in pure oxygen, but 20 in CO2 and 10 more in the water. (20 + 10 = 30). How do we get 30 in the pure oxygen?
Pentane combusts 2C5H10(G) + 15O2(G) 10CO2(G) + 10H2O(G) Change the pure oxygen to a 15O2 (15 x 2 = 30) now balanced.
SKELETON Rubidium metal is put into calcium nitrate solution _Rb(S) + _Ca(NO3)2 (AQ) _RbNO3(AQ) + _Ca(S) Rubidium makes a Rb+1 cation Nitrate is on table E, NO3-1 anion Calcium makes a Ca+2 cation Nitrate is on table E, NO3-1 anion Criss cross to get proper formula Criss cross to get proper formula
Rubidium metal is put into calcium nitrate solution _Rb(S) + _Ca(NO3)2 (AQ) 2RbNO3(AQ) +_ Ca(S) SAY: 1 Rb in reactants, and 1 Rb in products, so far so good. Then 1 Ca in reactants, and 1 a in products, also good. But two nitrates in the reactant side, so DOUBLE the RbNO3 to double nitrates in the products (then start over).
Rubidium metal is put into calcium nitrate solution 2Rb(S) + _Ca(NO3)2 (AQ) 2RbNO3(AQ) +_ Ca(S) SAY: 1 Rb in reactants, and 2 Rb in products, put a 2 in front of Rb in the reactants.
Rubidium metal is put into calcium nitrate solution 2Rb(S) + _Ca(NO3)2 (AQ) 2RbNO3(AQ) +_ Ca(S) SAY: Rb good 2:2 Ca good at 1:1 nitrates good at 2:2 Balanced! In a 2:1:2:1 ratio John Dalton would be happy, I m happy. You be happy too.
Rubidium metal is put into calcium nitrate solution 2Rb(S) + Ca(NO3)2 (AQ) 2RbNO3(AQ) + Ca(S) Why you might have got this one wrong: Rubidium makes only a +1 cation so forming rubidium nitrate can only be in a 1:1 ratio. The 2 in the calcium nitrate needs to be BALANCED away, not by you sticking a 2 in as in: Rb(NO3)2because it s easier to do. This: Rb(NO3)2 is just not possible.
Cobalt (III) chlorate solution and magnesium chromate solutions are mixed together
SKELETON Cobalt (III) chlorate solution and magnesium chromate solutions are mixed together _Co(ClO3)3 (AQ) + _MgCrO4(AQ) This is the set up for a double replacement reaction, we re starting with 2 AQ solutions. Don t just do this in your head, in our class we will make the switch, and balance the equation. If by chance both products are AQ also, that means it was NO REACTION. For practice balancing, we always go through the motions. You can t decide no reaction until it s balanced and THEN you go to table F to find the solid precipitate, if it s even there.
Cobalt (III) chlorate solution and magnesium chromate solutions are mixed together _Co(ClO3)3 (AQ) + _MgCrO4(AQ) _CoCrO4(?) + _MgClO3(?) Switch the anions around, here the chromate anion starts with the Mg, but ends up with the Co. The chlorate anion starts with Co, but ends up with the Mg. It s a double switch (like a square dance, switch your partners, do si do). FIX IN NEXT STEP! Note, we have NO IDEA YET about the solubility of the products, nor would we look before balancing the equation. Switch, fix, balance, and then table F them.
Cobalt (III) chlorate solution and magnesium chromate solutions are mixed together _Co(ClO3)3 (AQ) + _MgCrO4(AQ) _CoCrO4(?) + _MgClO3(?) We switched em. Now FIX EM! Co+3 and CrO4-2 gets criss-crossed to : Co2(CrO4)3 Mg+2 and ClO3-1 gets criss-crossed to: Mg(ClO3)2 _Co(ClO3)3 (AQ) + _MgCrO4(AQ) Co2(CrO4)3(?) + Mg(ClO3)2(?)
Cobalt (III) chlorate solution and magnesium chromate solutions are mixed together 2Co(ClO3)3 (AQ) + _MgCrO4(AQ) _Co2(CrO4)3(?) + _Mg(ClO3)2(?) We switched em, we fixed em now balance em! SAY: one Co in reactants, but two in products Put a 2 in front of Co(ClO3)3
Cobalt (III) chlorate solution and magnesium chromate solutions are mixed together 2Co(ClO3)3 (AQ) + _MgCrO4(AQ) _Co2(CrO4)3(?) + 3Mg(ClO3)2(?) SAY: 2 Co in reactants, and 2 in products good! SIX chlorates, in reactants, two in products, Put a 3 in front of Mg(ClO3)2 to get six chlorates in products, start over.
Cobalt (III) chlorate solution and magnesium chromate solutions are mixed together 2Co(ClO3)3 (AQ) + 3MgCrO4(AQ) _Co2(CrO4)3(?) + 3Mg(ClO3)2(?) SAY: 2 Co in reactants, and 2 in products good! SIX chlorates, in reactants, SIX in products .good! Only 1 Mg in reactants, but 3 are in the products, so put a 3 in front of MgCrO4 , start over again.
Cobalt (III) chlorate solution and magnesium chromate solutions are mixed together 2Co(ClO3)3 (AQ) + 3MgCrO4(AQ) _Co2(CrO4)3(?) + 3Mg(ClO3)2(?) SAY: 2 Co in reactants, and 2 in products good! SIX chlorates, in reactants, SIX in products .good! 3 Mg in reactants, and 3 are in the products good! 3 chromates in reactants, and LOOK, 3 chromates in products, balanced! This is a 2:3:1:3 ratio, it can t be reduced so - John Dalton is happy!
Cobalt (III) chlorate solution and magnesium chromate solutions are mixed together 2Co(ClO3)3 (AQ) + 3MgCrO4(AQ) _Co2(CrO4)3(S) + 3Mg(ClO3)2(AQ) LAST STEP: check products on table F, which is AQ or S??? If your last name is chromate (right side) you are insoluble but there are exceptions like Mg, but not Co, so the first product is SOLID! A precipitate is the proof that a DR happened! If your last name is chlorate, you are soluable (AQ) without exception! DONE finally!
SKELETON Hydrogen monoiodide gas decomposes _HI(G) _H2(G) + _I2(S) Hydrogen mono-ioidide is a 1:1 ratio, and the both products are HONClBrIF twins, diatomic. The reactant is a gas, we re told that. Hydrogen is a gas normally, and iodine is a solid at normal temperatures.
SKELETON Hydrogen monoiodide gas decomposes 2HI(G) _H2(G) + _I2(S) SAY: One H in reactant, two in products. Put a 2 in front of HI, and start over.
Hydrogen monoiodide gas decomposes 2HI(G) _H2(G) + _I2(S) SAY: 2 H s in reactants, and two in products, good. SAY: 2 I s in reactants, and also in products. Balanced! (this one was easy)
SKELETON Carbon and hydrogen form ethane (C2H6) gas _C(S) + _H2(G) _C2H6(G) Carbon is a solid and it is atomic, just C Hydrogen is a gas, and is diatomic HONClBrIF Twin You re given the formula for ethane and it s a gas.
Carbon and hydrogen form ethane (C2H6) gas 2C(S) + _H2(G) _C2H6(G) SAY: one C in reactant, but two in products, put a 2 in front of carbon, start over.
Carbon and hydrogen form ethane (C2H6) gas 2C(S) + 3H2(G) _C2H6(G) SAY: two C in reactants, and two in products, good. SAY: 2 H s in reactants, but 6 H s in products. How to get six H s in front? Add a three in front of the pure hydrogen. Start over again!
Carbon and hydrogen form ethane (C2H6) gas 2C(S) + 3H2(G) _C2H6(G) SAY: two C in reactants, and two in products, good. SAY: 6 H s in reactants, but 6 H s in products. Good! Balanced, this was easier also. In a 2:3:1 ratio (JD says yay!
Propanol combusts (C3H7OH) Propanol is a type of alcohol, it s an oxygenated hydrocarbon. That means it is made up of hydrogen and carbon (like a hydrocarbon) PLUS some oxygen. This is fine, but it makes balancing a bit trickier, as now there is oxygen in both reactants as well as both products, but you re smart, this is fun.
SKELETON Propanol combusts (C3H7OH) _C3H7OH(G) + _O2(G) _CO2(G) + _H2O(G) In combustion, a hydrocarbon, or OXYGENTATED HYDROCARBON combines rapidly with oxygen and forms only carbon dioxide and water. Pure oxygen is diatomic, and the product formulas are well known already.
Propanol combusts (C3H7OH) _C3H7OH(G) + _O2(G) 3CO2(G) + _H2O(G) SAY: 3 carbons in the reactants, so put a 3 in front of CO2, start over.
Propanol combusts (C3H7OH) _C3H7OH(G) + _O2(G) 3CO2(G) + _H2O(G) SAY: 3 carbons in the reactants, and 3 in products, good.
Propanol combusts (C3H7OH) _C3H7OH(G) + _O2(G) 3CO2(G) + 4H2O(G) SAY: 3 carbons in the reactants, and 3 in products, good. SAY: 7 + 1 H = 8H in reactants, we need to put a 4 in front of the water. Start over.
Propanol combusts (C3H7OH) _C3H7OH(G) + _O2(G) 3CO2(G) + 4H2O(G) SAY: 3 carbons in the reactants, and 3 in products, good. SAY: 7 + 1 H = 8H in reactants, and 8 H in products, good. SAY: two oxygens in reactants, but 6 + 4 = 10 oxygen in products (NOT GOOD). To fix this, think we need even oxygen on reactant side, we need to double the propanol to do that, then start over.
Propanol combusts (C3H7OH) 2C3H7OH(G) + _O2(G) 3CO2(G) + 4H2O(G) SAY: 6 carbons in the reactants, and 3 in products, to fix this, change the 3 into a 6, and start over.
