# Calculate the volume and mass of carbon monoxide (carbon monoxide (2)) required to reduce copper from copper oxide

**Calculate the volume and mass of carbon monoxide (carbon monoxide (2)) required to reduce copper from copper oxide (2) weighing 32 g, containing 15% impurities.**

1. Let’s find the mass of pure copper. If CuO contains 15% of impurities, then the pure substance will be 85%.

32 g – 100%

x g – 85%

Hence, x = 32 * 85/100 = 27.2 g CuO

2. Let’s compose the reaction equation:

CuO + CO = Cu + CO2

According to the equation, all substances are taken in an amount of 1 mol.

Let’s find the mass of CuO by the formula:

m (CuO) = n * M = 1 * (64 + 16) = 80 g

Let’s find the mass of CO by the formula:

m (CO) = n * M = 1 * (12 + 16) = 28 g

3. Let’s calculate the mass of CO, making up the proportion:

27.2 g CuO – x g CO

80 g CuO – 28 g CO

Hence, x = 27.2 * 28/80 = 9.52 g

4. Let’s calculate the volume of CO by the formula:

V (CO) = Vm * m / M = 22.4 * 9.52 / 28 = 7.6 l