5 Stoichiometry Exercises With Answers to Try Now

Stoichiometry is a fundamental concept in chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It is a crucial tool for chemists to predict the amount of substances required or produced in a reaction. In this article, we will go through five stoichiometry exercises with answers to help you practice and reinforce your understanding of this concept.

Understanding Stoichiometry Basics

Before we dive into the exercises, let’s quickly review some key concepts in stoichiometry:

• Mole ratio: The ratio of the number of moles of one substance to the number of moles of another substance in a balanced chemical equation.
• Limiting reactant: The reactant that is consumed first in a reaction, determining the amount of product formed.
• Excess reactant: The reactant that is left over after the limiting reactant is consumed.

Exercise 1: Mole Ratio Calculation

Calculate the mole ratio of carbon dioxide (CO2) to glucose (C6H12O6) in the following balanced equation:

C6H12O6 + 6O2 → 6CO2 + 6H2O

Answer: The mole ratio of CO2 to C6H12O6 is 6:1.

Exercise 2: Limiting Reactant Determination

Consider the following reaction:

2Al + 3CuSO4 → Al2(SO4)3 + 3Cu

If 2 moles of Al and 4 moles of CuSO4 are present, which reactant is the limiting reactant?

Answer: Al is the limiting reactant because 2 moles of Al would require 3 moles of CuSO4, but only 2 moles of CuSO4 are present.

Exercise 3: Mass-Mass Calculation

Calculate the mass of copper (Cu) produced in the following reaction:

2Al + 3CuSO4 → Al2(SO4)3 + 3Cu

If 25 grams of Al and 50 grams of CuSO4 are used, what is the mass of Cu produced?

Answer:

First, calculate the number of moles of Al and CuSO4:

moles Al = 25 g / 27 g/mol = 0.93 mol moles CuSO4 = 50 g / 160 g/mol = 0.31 mol

Next, determine the limiting reactant (Al) and calculate the number of moles of Cu produced:

moles Cu = 0.93 mol Al x (3 mol Cu / 2 mol Al) = 1.4 mol

Finally, calculate the mass of Cu produced:

mass Cu = 1.4 mol x 63.5 g/mol = 89 g

Exercise 4: Mole-Mole Calculation

Calculate the number of moles of oxygen (O2) required to produce 2 moles of carbon dioxide (CO2) in the following reaction:

C6H12O6 + 6O2 → 6CO2 + 6H2O

Answer:

First, identify the mole ratio of O2 to CO2:

6 moles O2 : 6 moles CO2

Next, calculate the number of moles of O2 required:

moles O2 = 2 moles CO2 x (6 mol O2 / 6 mol CO2) = 2 mol

Exercise 5: Empirical Formula Calculation

A compound has the following percentage composition:

Calculate the empirical formula of the compound.

Answer:

First, assume a 100-gram sample and calculate the mass of each element:

mass C = 40 g mass H = 6.7 g mass O = 53.3 g

Next, calculate the number of moles of each element:

moles C = 40 g / 12 g/mol = 3.33 mol moles H = 6.7 g / 1 g/mol = 6.7 mol moles O = 53.3 g / 16 g/mol = 3.33 mol

Finally, divide the number of moles of each element by the smallest number of moles (3.33 mol):

C: 3.33 mol / 3.33 mol = 1 H: 6.7 mol / 3.33 mol = 2 O: 3.33 mol / 3.33 mol = 1

The empirical formula is CH2O.

In conclusion, these exercises demonstrate the importance of understanding stoichiometry in chemistry. By mastering mole ratios, limiting reactants, and empirical formulas, you can accurately predict the amount of substances required or produced in chemical reactions.

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