Stoichiometry : Mole Ratios

Stoichiometry is the study of quantitative relationships. A balanced chemical equation is required. Mole ratios are the first part of any stoichiometric problem. They are used to determine the relationship between the known quantity and the unknown quantity. The setup is as follows:

mol unknown/mol known

Coefficients are used to relate the quantities. This quiz will cover proper identification of required mole ratios from chemical equations. Select the best answer from the choices.

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C3H8 + 502 --> 3CO2 + 4H2O If there are two moles of oxygen available, which mole ratio will tell you how much water is produced?

4 mol H20/5 mol O2

3 mol H20/5 mol O2

4 mol O2/4 mol H2O

5 mol O2/4 mol H2O

C3H8 + 502 --> 3CO2 + 4H2O If there are 10 moles of C3H8, which mole ratio will tell you how much oxygen is required for the reaction to take place?

1 mol C3H8/5 mol O2

4 mol C3H8/5 mol O2

5 mol O2/1 mol C3H8

3 mol O2/1 mol C3H8

C3H8 + 502 --> 3CO2 + 4H2O Which ratio will tell you how much water is produced from the combustion of .5 moles of C3H8?

1 mol C3H8/4 mol H2O

4 mol H2O/1 mol C3H8

3 mol H2O/1 mol C3H8

6 mol C3H8/4 mol H2O

C3H8 + 502 --> 3CO2 + 4H2O Classify the above reaction.

Synthesis

Composition

Decomposition

None of these

4Al + 3O2 --> 2Al2O3 Which mole ratio tells you how much product is produced from three moles of aluminum?

4 mol Al/2 mol Al2O3

5 mol Al2O3/4 mol Al

4 mol Al2O3/4 mol Al

2 mol Al2O3/4 mol Al

4Al + 3O2 --> 2Al2O3 Which mole ratio describes how much oxygen is required to produce 10 moles of product?

4 mol Al2O3/3 mol O2

9 mol O2/2 mol Al2O3

3 mol O2/2 mol Al2O3

2 mol Al2O3/3 mol O2

4Al + 3O2 --> 2Al2O3 Classify the above reaction.

Synthesis

Combustion

Decomposition

Generation

ZnO + 2HCl --> ZnCl2 + H2O Which mole ratio tells you have to find how much HCl is required to get 8 moles of ZnCl2?

4 mol HCl/1 mol ZnCl2

2 mol HCl/3 mol ZnCl2

2 mol HCl/1 mol ZnCl2

1 mol ZnCl2/2 mol HCl

Quiz/Test Summary
Title: Stoichiometry : Mole Ratios
Questions: 8
Contributed by:
Steve