Question

1. _2_ cu + _1_ o₂ → _2_ cuo

2a if 101 grams of copper is used, how many moles of copper (ii) oxide will be formed?
2b if 78.2 grams of oxygen react with excess copper, how many moles of copper (ii) oxide will be produced?
2c what mass of copper is used to produce 2.87 moles of copper (ii) oxide?

Explanation:

2a

Step1: Find moles of Cu

Molar mass of Cu is 63.55 g/mol. Moles of Cu = mass / molar mass = \( \frac{101\ g}{63.55\ g/mol} \approx 1.59\ mol \).

Step2: Use stoichiometry

From the equation \( 2Cu + O_2
ightarrow 2CuO \), the mole ratio of Cu to CuO is 2:2 (or 1:1). So moles of CuO = moles of Cu ≈ 1.59 mol.

Step1: Find moles of \( O_2 \)

Molar mass of \( O_2 \) is 32.00 g/mol. Moles of \( O_2 \) = \( \frac{78.2\ g}{32.00\ g/mol} \approx 2.444\ mol \).

Step2: Use stoichiometry

From the equation, mole ratio of \( O_2 \) to CuO is 1:2. So moles of CuO = 2 × moles of \( O_2 \) = 2 × 2.444 mol ≈ 4.888 mol.

Step1: Use stoichiometry for Cu to CuO

Mole ratio of Cu to CuO is 1:1 (from \( 2Cu + O_2
ightarrow 2CuO \)). So moles of Cu = moles of CuO = 2.87 mol.

Step2: Find mass of Cu

Mass of Cu = moles × molar mass = 2.87 mol × 63.55 g/mol ≈ 182 g.

Answer:

Approximately 1.59 moles

2b