Answer to Question #58271 in General Chemistry for sara

Question

f 13.2kg of Al reacts with an excess of Fe2O3, how many kg of Al2O3 will be produced?

Answer 1

Fe2O3 + 2Al → Al2O3 + 2Fe
M(Fe2O3) = 159.7 kg/kmol
n(Fe2O3) = 13.2kg/159.7kg/kmol = 0.083 kmol
M(Al2O3) = 102 kg/kmol
n(Al2O3) = 2n(Fe2O3) = 2∙0.083 kmol = 0.165 kmol
m(Al2O3) = 0.165kmol∙102 kg/kmol = 16.83 kg.
Answer: 16.83 kg.