Question #58271

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

Expert's answer

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.

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.

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