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6th Hour
Stoichiometry Review
Key Terms:
composition stoichiometry reaction stoichiometry
mole ratio limiting reactant excess reactant
theoretical yield actual yield percentage yield
Key Skills:
balance an unbalanced equation determine the molar mass of a compound identify the mole ratio between two substances in a chemical reaction perform the following ideal stoichiometric calculations o mole-to-mole o mole-to-mass o mass-to-mole o mass-to-mass determine the limiting reactant of a chemical reaction determine the percentage yield given an amount of actual yield
Practice Problems:
- Balance the unbalanced equation below, and then determine the following molar ratios. 2 Al (^) ( s ) +3 H 2 SO4 ( aq ) → 1 Al 2 (SO 4 )3 ( aq ) + 3 H2 (g) a. H 2 SO 4 to Al
b. Al 2 (SO 4 ) 3 to Al
c. H 2 to H 2 SO 4
d. Al to H 2 SO 4
- Carborundum, SiC, is a hard substance made by combining silicon dioxide with carbon coke as described by the unbalanced equation below. What mass in grams of SiC is formed from the complete reaction of 2. mol of carbon?
1 SiO2 ( s ) + 3 C (^) ( s ) → 1 SiC (^) ( s ) + 2 CO (^) ( g )
2.00 mol C 1 mol SiC 40.10 g SiC 26.7 g SiC 3 mol C 1 mol SiC
6th Hour
Stoichiometry Review
6th Hour
Stoichiometry Review
- 427.5 g of iron (III) hydroxide is combined with 637.5 g of sulfuric acid to produce iron (III) sulfate and water.
2 Fe(OH)3 ( s ) + 3H 2 SO4 ( aq ) → Fe 2 (SO 4 )3 ( s ) + 6 H 2 O (^) ( l )
a. What is the limiting reactant?
427.5 g Fe(OH) 3 1 mol Fe(OH) 3 3 mol H 2 SO 4 98.08 g H 2 SO 4 588.5 g H 2 SO 4 106.88 g Fe(OH) 3 2 mol Fe(OH) 3 1 mol H 2 SO 4
588.5 g H 2 SO 4 are required to fully react 427.5 g Fe(OH) 3. Since 637.5 g of H 2 SO 4 is available it is the excess reactant, making Fe(OH) 3 the limiting reactant.
b. What mass of excess react remains once the reaction is complete?
637.5 g H 2 SO 4 - 588.5 g H 2 SO 4 = 49.0 g H 2 SO 4
49.0 g of H 2 SO 4 remains
c. What mass of iron (III) sulfate is produced?
427.5 g Fe(OH) 3 1 mol Fe(OH) 3 1 mol Fe 2 (SO 4 ) 3 399.88 g Fe 2 (SO 4 ) 3 799.7 g Fe 2 (SO 4 ) 3 106.88 g Fe(OH) 3 2 mol Fe(OH) 3 1 mol Fe 2 (SO 4 ) 3
- If 32.0 g of calcium carbide (CaC 2 ) reacts as described by the unbalanced reaction below, and 11.3 g of acetylene gas (C 2 H 2 ) is produced. What is the percentage yield?
1 CaC 2 + 2 H 2 O → 1 C 2 H 2 + 1 Ca(OH) 2
Theoretical yield of C 2 H 2 :
32.0 g CaC 2 1 mol CaC 2 1 mol C 2 H 2 26.04 g C 2 H 2 13.0 g C 2 H 2 64.10 g CaC 2 1 mol CaC 2 1 mol C 2 H 2
