Question

Chromium metal reacts with oxygen to give chromium(III) oxide.

A. Write a balanced Chemical equation for the reaction.
B. What mass (in g) of chromium(III) oxide is produced if 0.250 g of chromium metal is converted completely to the oxide?
C. What mass of oxygen (in g) is required for the reaction?
D. If 0.098 g of oxygen was isolated, what was the percent yield?

Answer 1

The reaction between chromium and oxygen forms chromium(III) oxide. To find the mass of the oxide produced and oxygen needed, use stoichiometry based on the balanced equation and molar masses. The percent yield is calculated from the actual and theoretical mass of oxygen involved.

Step-by-step explanation:

Chromium Reaction with Oxygen

The reaction between chromium metal and oxygen to form chromium(III) oxide can be represented by the following balanced chemical equation:

4 Cr(s) + 3 O2(g) → 2 Cr2O3(s)

To calculate the mass of chromium(III) oxide produced from 0.250 g of chromium:

1. Determine the molar mass of Cr and Cr2O3 (51.996 g/mol for Cr, and 151.99 g/mol for Cr2O3).
2. Calculate mols of Cr (0.250 g divided by 51.996 g/mol).
3. Use the stoichiometry of the balanced equation (4 mol Cr produces 2 mol Cr2O3) to find mols of Cr2O3 produced.
4. Convert mols of Cr2O3 to grams (mols multiplied by molar mass of Cr2O3).

To find the mass of oxygen required:

1. Use the stoichiometry of the balanced equation to relate mols of Cr to mols of O2 (4 mol Cr to 3 mol O2).
2. Determine the molar mass of O2 (32.00 g/mol).
3. Calculate the mass of O2 by multiplying the mols of O2 needed by its molar mass.

If 0.098 g of oxygen was actually used, the percent yield is found using the actual mass of oxygen obtained divided by the theoretical mass required, multiplied by 100%.