A mixture of CuO and Cu2O with a mass of 10.50 g is reduced to give 8.66 g of pure Cu metal. What are the amounts in grams of CuO and Cu2O in the original mixture?

Name: Chemistry
Author: McMurry
ISBN: 9781292336145

Verified step by step guidance

Define the chemical reactions involved: CuO + H2 -> Cu + H2O and Cu2O + H2 -> 2Cu + H2O.

Let x be the mass of CuO and y be the mass of Cu2O in the mixture. Therefore, x + y = 10.50 g.

Use the molar masses: CuO (79.55 g/mol), Cu2O (143.1 g/mol), and Cu (63.55 g/mol) to set up equations based on the mass of Cu produced.

From the reaction stoichiometry, CuO produces 1 mole of Cu per mole of CuO, and Cu2O produces 2 moles of Cu per mole of Cu2O. Set up the equation: (x/79.55) * 63.55 + (y/143.1) * 2 * 63.55 = 8.66 g.

Solve the system of equations: x + y = 10.50 and the equation from step 4 to find the values of x and y.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the reactants and products in a chemical reaction. It allows us to calculate the amounts of substances involved based on balanced chemical equations. In this problem, stoichiometry will help determine how much CuO and Cu2O were present in the original mixture by relating their masses to the mass of the produced Cu.

Reduction Reactions

Reduction reactions involve the gain of electrons or a decrease in oxidation state by a substance. In this context, CuO and Cu2O are reduced to form elemental copper (Cu). Understanding the reduction process and the oxidation states of copper in these compounds is essential for determining how much of each compound was present in the original mixture.

Mass Conservation

The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. This principle is crucial for solving the problem, as the total mass of the reactants (CuO and Cu2O) must equal the mass of the products (pure Cu and any remaining oxygen). By applying this law, we can set up equations to find the original amounts of CuO and Cu2O in the mixture.

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