I have no idea how to even start, I have already tried this problem several time

Question

I have no idea how to even start, I have already tried this problem several times and still can't get it. Thank you in advance! What is the percent yield in a reaction between 52.4 g O2 and 56.3 g Al if 99.1 g of Al2O3 is produced?

Explanation / Answer

2Al + 1.5O2 ---> Al2O3 56.3 gm Al = 56.3/26.98 = 2.086 moles 52.4 gm O2 = 52.4/32 = 1.6375 99.1 gm Al2O3 produced 2.086 moles Al gives 2.086/2 =1.043 moles Al2O3 = 1.043 x 101.96 = 106.345 gm is expected % yield = expected mass of alumina/actyal mass produces x 100 *% yiles = (99.1/106.345) x100 = 93.18 %

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