Question

A laboratory test of 11.8 grams of copper ore yields 5.24 grams of copper. If the copper compound in the ore is CuO and it is converted to the pure metal by the reaction 2 CuO(s) + C(s) 2 Cu(s) + CO2(g), what is the percentage of CuO in the ore?

Answer 1

Molar mass of Cu = 63.55 g/mol

mass of Cu = 5.24 g
molar mass of Cu = 63.55 g/mol
mol of Cu = (mass)/(molar mass)
= 5.24/63.55
= 0.0825 mol


Balanced chemical equation is:
2 CuO(s) + C(s) —> 2 Cu(s) + CO2(g)


According to balanced equation
mol of CuO reacted = moles of Cu
= 0.0825 mol



mass of CuO = number of mol * molar mass
= 0.0825*79.55
= 6.56 g

This is mass of CuO in ore

% CuO = mass of CuO * 100 / mass of Ore
= 6.56 * 100 / 11.8
= 55.6 %

Answer: 55.6 %