MCuO = 80kg/kmol
n = m/M = 814/80 = 10,175 kmoli CuO
n = m/M = 62/12 = 5,166 kmoli C
a) n CuO exces = 10,175-5,1666 = 5 kmoli
m CuO = 5×80 = 400,66kg
b) 2CuO + C => 2Cu + CO2
1 kmol C..........2×64kg Cu
5,166 kmoli..........x = 661,3248kg Cu
c) n C = n CO2 = 5,166 kmoli
Ca(OH)2 + CO2 => CaCO3 + H2O
n Ca(OH)2 = nCO2 = 5,166 kmoli
md Cu(OH)2 = 5,166 ×74 = 382,284kg
ms = md×100/C = 38228,4 kg sol Ca(OH)2
