Determine the percent yield of a reaction that produces 28 65 g of Fe when 50.0

Question

Determine the percent yield of a reaction that produces 28 65 g of Fe when 50.0 00 g of Fez0s react with excess Al according to the following reaction Fe2O3(s)+2 Al(s) Al2O3(s)+2 Fe(s)

W-SUMMERZC xCourse Home urse html?courseld-14733194&OpenVellumHMAC; 01b27od2b2d7819a51616320b79deb92 10001 CHIM 151 0W Summer 2018 SCC e Home xam 3 2 Hide em Constant Part A Determine the percent yield of a reaction that produces 28 65 g of Fe when 50 00 g of Fe0s react with excess Al according to the following reaction Foz0sls)+ 2 AS)Al0s)+2 Fe(s) 8193% 20 02 % -57 30 % o 2865 % 61 03 Submit Provide Feedback 5 6

Explanation / Answer

The answer is a ( 81.9%).

moles of Fe2O3 = weight in gram / molecular mass = 50 / 159.69 = 0.313 moles

According to the given reaction ; 1 mole of Fe2O3 produce 2 moles of Fe

so, 0.313 moles of fe2O3 will produce = 0.313 x 2 = 0.626 moles of Fe

weight of Fe in grams:

moles = weight in gram / molecular mass

weight in grams = 0.626 x 55.847 = 34.97 gram Fe (theoritical yield)

% yield = (Actual yield / theoritical yield ) x 100

% yield = (28.65 / 34.97) x 100 = 0.819 x 100 = 81.9 %

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