A mixture contains Al2O3 (s) [MW=101.96 g/mol] and Fe2O3 (s) [MW=159.69 g/mol] a

Question

A mixture contains Al2O3 (s) [MW=101.96 g/mol] and Fe2O3 (s) [MW=159.69 g/mol] and weighs 1.983g. when heated under excess hydrogen gas, the iron oxide reacts to form water vapor and Fe(s) but the aluminum oxide remains unreacted. if the residue following heating weighs 1.596g, determine the weight percent of iron oxide in the original mixture.


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Explanation / Answer

Step1 Let mass of Al2O3 = x ; Mass of Fe2O3= 1.983 - x Step2 Fe2O3(s) +3 H2(g) ----> 2 Fe(s) + 3 H2O(l) Step3 Moles of Fe2O3 = (1.983 -x)/159.69 Step4 Moles of Fe = 2 (1.983-x)/159.69 Step5 Mass of Fe = 2 (1.983-x) x111.69/159.69 Step6 x + 2(1.983-x) * 111.69/159.69 = 1.596 Step7 Solve for x.

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