STOICHIOMETRY: tells relative amts of reactants &
products in a chemical reaction
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To solve stoichiometry problems… ALWAYS!!!!!!!!!!!
** WRITE THE BALANCED
EQN & GIVEN INFORMATION! **
1.) Find moles of given element or compound.
*
Use molar mass of given substance, if problem gives you grams.
2.) Use mole ratio (coefficients) from balanced
equation.
3.) Find answer.
* Use molar mass of unknown substance,
if question asks for grams.
So… here’s an example:
If 34.32 grams of acetylene (C2H2) are burned in air, how
many moles of CO2 can be formed?
2 C2H2 + 5 O2
à 4 CO2 + 2 H2O
34.32 g
? moles
1.) 34.32 g C2H2 | 1 mole C2H2
= 1.32 moles C2H2 C: 2 x 12.0 = 24.0
|
26 grams H:
2 x 1.0 = 2.0 +
26.0
2.) 1.32 moles C2H2 = x moles
CO2 2
x = 5.28
2 4
x = 2.64 moles CO2
Problem
asked for moles of CO2, so that is your answer. Be sure to round for significant figures.
Answer should be reported as 2.640 moles
CO2
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Example
2:
How
many grams of Na2CO3 are needed to react completely with
6.21 moles of Ca(OH)2?
Na2CO3 +
Ca(OH)2 à 2 NaOH + CaCO3
? g
6.21 moles
1.) Step 1 is already done for you!
The number of moles is given in the problem! YAY!
2.) 6.21 moles Ca(OH)2
= x moles Na2CO3
x = 6.21 moles Na2CO3
1 1
3.) 6.21 moles Na2CO3 | 106 grams = 658 g Na2CO3
| 1 mole
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Example 3:
If
27.8 grams of Na2SO4 are reacted with excess Al2(CO3)3, how many grams
of Al2(SO4)3 will be formed?
3
Na2SO4 + Al2(CO3)3 à Al2(SO4)3 + 3 Na2CO3
27.8 g ? g
1.) 27.8 g Na2SO4
| 1 mole Na2SO4 = 0.196 moles Na2SO4 Na: 2 x 23.0 = 46.0
|
142.1 grams S: 1 x 32.1 = 32.1
O:
4 x 16.0 = 64.0
+
2.) 0.196 moles Na2SO4 = x mole Al2(SO4)3 142.1
3 1 3
x = 0.196
x = 0.0653 moles Al2(SO4)3
3.) 0.0653
mole Al2(SO4)3 | 342.3 g Al2(SO4)3
= 22.4 g Al2(SO4)3 Al:
2 x 27.0 = 54.0 | 1 mole Al2(SO4)3 S: 3 x 32.1
= 96.3
O:
12 x 16.0 = 192.0 + 342.3
Practice
Problem:
How
many grams of calcium nitrate are formed when 57.9 grams of iron (III) nitrate
react with excess calcium hydroxide according to the following equation?
___ Ca(OH)2 + ___
Fe(NO3)3 à ___ Ca(NO3)2 + ___
Fe(OH)3