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(MasstoMass calculation) What mass of Al is needed to reduce 10.0 kg of Cr(III) oxide to produce chromium metal? 2Al(l)+Cr_(2)O_(3)(s) overset(Delta)to Al_(2)O_(3)(s)+2Cr(l) |
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Answer» Solution :`M(Cr_(2)O_(3))=152.0" g "mol^(-1),1 " mol "Cr_(2)O_(3)=2" mol "Al,M_(Al)=26.98" g "mol^(-1)` Now apply the formulas: `n_(Cr_(2)O_(3))=(10.0xx10^(3)g)/(152" g "mol^(-1))=65.79mol,n_(Al)=2xxn_(Cr_(2)O_(3))=2xx65.78mol=131.58` mol `THEREFORE m_(Al)=n_(Al)*M_(Al)=131.58molxx26.98g" "mol^(-1)=3.550g=3.55`KG |
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