Calculate the volume of oxygen that was consumed in the oxidation reaction of a portion of copper

Calculate the volume of oxygen that was consumed in the oxidation reaction of a portion of copper a) weighing 128 g b) volume 57.6 cm3 (the density of copper is taken equal to 9 g / cm3) For both cases, calculate the mass of copper oxide, which forms as a result of the reaction.

Copper oxidation is described by the equation:

2Cu + O2 = 2CuO;

Based on condition a):

1. find the chemical amount of copper:

n (Cu) = m (Cu): M (Cu) = 128: 64 = 2 mol;

2. according to the reaction equation, the amount of oxygen is two times less:

n (O2) = n (Cu): 2 = 1 mol;

3. find the volume of oxygen:

V (O2) = n (O2) * Vm = 1 * 22.4 = 22.4 dm3;

4.Calculate the mass of copper oxide:

n (CuO) = n (Cu) = 2 mol;

m (CuO) = n (CuO) * M (CuO) = 2 * 80 = 160 g.

Based on condition b):

1.Calculate the mass of a portion of copper:

m (Cu) = ρ * V = 9 * 57.6 = 518.4 g;

n (Cu) = 518.4: 64 = 8.1 mol;

2.find the volume of oxygen:

n (O2) = 8.1: 2 = 4.05 mol;

V (O2) = 4.05 * 22.4 = 90.72 dm3;

3. find the mass of copper oxide:

n (CuO) = 8.1 mol;

m (CuO) = 8.1 * 80 = 648 g.

Answer: a) 22.4 dm3, 160 g; b) 90.72 dm3, 648 g.