Calculate the volume of carbon monoxide 2 which will require 8 kg of iron oxide 3 for reduction to metal

Calculate the volume of carbon monoxide 2  which will require 8 kg of iron oxide 3 for reduction to metal, which mass is necessary to obtain the required amount of carbon monoxide 2.

Given:
m (Fe2O3) = 8 kg = 8000 g

To find:
V (CO) -?
m (CO) -?

1) Write the reaction equation:
3CO + Fe2O3 => 2Fe + 3CO2;
2) Calculate the amount of the substance Fe2O3:
n (Fe2O3) = m (Fe2O3) / M (Fe2O3) = 8000/160 = 50 mol;
3) Determine the amount of CO:
n (CO) = n (Fe2O3) * 3 = 50 * 3 = 150 mol;
4) Calculate the volume of CO (under normal conditions):
V (CO) = n (CO) * Vm = 150 * 22.4 = 3360 l;
5) Calculate the mass of CO:
m (CO) = n (CO) * M (CO) = 150 * 28 = 4200 g.

Answer: The volume of CO is 3360 liters; weight – 4200 g.