Vocabulary
Mole-Mole Conversions
Mass-Mass Converstions
Limiting Reactant/Percent Yield
Conversions Using Volume
100

___________is the use of the relationships in a balanced chemical equation to carry out calculations involving the reactants and products.

Stoichiometry

100

Consider the following chemical equation:

2 N2H4(g) + N2O4(g) → 3 N2(g) + 4 H2O(g)

How many moles of water form from 2.0 moles of N2O4?

8.0 mol H2O

100

How much oxygen (in grams) will form from 23.3 g of N2O5?

2 N2O5(s) + 4 NO2(g) → O2(g)

3.45 g O2

100

You react 10.0 g of N2 and 5.0 g of H2 in a flask. What is the limiting reactant?

N2(g) + 3 H2(g) → 2 NH3(g)

LR: N2

(Theoretical Yield = 12.2 g NH3)

100

How many milliliters of carbon dioxide gas at STP are produced from the decomposition of 8.57 g of iron (III) carbonate?

Fe2(CO3)3 (s) → Fe2O3 (s) + 3 CO2 (g)

1970 mL CO2 OR

1.97*103 mL CO2

200

When you perform a laboratory experiment, the amount of product collected is the __________________ yield.

Actual

200

Consider the following chemical equation:

2 N2H4(g) + N2O4(g) → 3 N2(g) + 4 H2O(g)

How many moles of nitrogen form from 2.0 moles of N2H4?

3.0 mol N2

200

How much methanol (CH3OH, in grams) can be formed from 65.4 g of hydrogen?

CO(g) + 2 H2(g) → CH3OH(g)

520 g CH3OH

200

Determine the mass (in g) of KOH formed if 10.0 g of KO2 reacts with 10.0 g of H2O and identify the limiting reactant. 

4 KO2(s) + 2 H2O(l) → 3 O2(g) + 4 KOH(s)

7.89 g KOH (Theoretical Yield)

LR: KO2

200

How many grams of mercury (II) oxide decompose to give 50.0 mL of oxygen gas at STP?

2 HgO (s) → 2 Hg (l) + O2 (g)

0.967 g HgO

300

The amount of product calculated from a limiting reactant problem is the __________________ yield.

Theoretical

300

Consider the following chemical equation:

2 H2S(g) + SO2(g) → 3 S(s) + 2 H2O(g)

How many moles of H2S must react to form 4.0 moles of S?

2.7 mol H2S

300

What mass in grams of NO2 is needed to form 197.4 g of NO?

NO2(g) + H2O(l) → HNO3(aq) + NO(g)

908.0 g NO2

300

Determine the mass (in g) of CuO formed from the reaction of 92.0 g Cu2O and 12.0 g O2 and identify the limiting reactant. 

Cu2O(s) + O2(g) → CuO(s)

102 g CuO (Theoretical Yield)

LR: Cu2O

300

What mass of copper (in grams) will form when 125.0 mL of CuSO4 reacts with excess aluminum?

2 Al(s) + 3 CuSO4(aq) → 3 Cu(s) + Al2(SO4)3(aq)

0.3546 g Cu

400

Chemists often determine the efficiency of a chemical reaction by calculating ___________ ________, which tells what percentage the actual yield is of the theoretical yield.

Percent Yield

400

Consider the following chemical equation:

C2H5OH(l) + O2(g) → CO2(g) + H2O(l)

How many moles of of CO2 form from the reaction of 3.33 moles of C2H5OH?

6.66 mol CO2

400

Determine the mass (in grams) of HBO2 formed from 40.0 g of B2H6

B2H6(g) + O2(g) → HBO2(g) + H2O(l)

127 g HBO2

400

When 49.8 g of NH3 reacts with 49.8 g of F2, what is the theoretical yield of HF? What is the percent yield if 7.22 g of HF is actually formed?

2 NH3(g) + 5 F2(g) → N2F4(g) + 6 HF(g)

31.5 g HF (Theoretical Yield)

22.9% Yield

400

Determine the mass (in g) of precipitate that forms when 78.8 mL of Na2CO3 reacts with excess calcium hydroxide.

Ca(OH)2(aq) + Na2CO3(aq) → NaOH(aq) + CaCO3(s)

0.352 g CaCO3

500

________________ ____________ are those present in quantities greater than necessary to react with the quantity of the limiting reactant.

Excess Reactants

500

Consider the following chemical equation:

C2H5OH(l) + O2(g) → CO2(g) + H2O(l)

How many moles of CO2 form when 2.94 moles of water form?

1.96 mol CO2

500

What mass in grams of C2H2 is needed to form 29.4 g of CO2?

C2H2(g) + O2(g) → CO2(g) + H2O(g)

8.70 g C2H2

500

Determine the limiting reactant, the mass of water formed (in g), and the mass that remains of the excess reactant when 20.0 g of HCl reacts with 20.0 g of O2.

HCl(g) + O2(g) → Cl2(g) + H2O(g)

LR: HCl

4.94 g H2O (Theoretical Yield)

15.6 Oleft over (in excess)

500

What volume (in mL) of NiCl2 solution is required to form 78.8 g of precipitate?

NiCl2(aq) + Na3PO4(aq) → Ni3(PO4)2(s) + NaCl(aq)

14500 mL NiCl2 OR

1.45*104 mL NiCl2

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