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
