Aluminum oxide was prepared by heating 225 g of chromium (III) oxide with 124 g of aluminum. Calculate the percent yield if 100.0 g of aluminum oxide were obtained

59.287%

Explanation:

Cr2O3+2Al -> Al2O3

this is your balanced equation

you are given mass of the chromium oxide (225g) and the aluminum (124g)

set up a limiting reactant problem and find how much Alumium oxide could be produced from each reactant.

chromiun oxide

225gCr2O3 * 1molCr2O3/151.99gCr2O3 * 1 molAl2O3/1molCr2O3 * 113.97gAl2O3 =168.67g Al2O3

now with aluminum

124gAl * 1molAl/26.98gAl * 1molAl2O3/2molAl *113.97gAl2O3/1molAl2O3 = 523.69g Al2O3

The smaller answer is your theoritical yield, so chromium oxide is your limiting reactant because it will make less aluminum oxide.

the equation to find percent yield=(actual/theoretical)*100

your actual yield is the 100g you actually made

So percent yield=(100.0g/168.67)*100

=59.287

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the answer is c. nacl