A 1.25-g sample contains some of the very reactive compound Al(C6H5)3. On treating the compound with aqueous HCl, 0.951 g of C6H6 is obtained.

Question

Al(C6H5)3(s) + 3HCl(aq) --> AlCl3(aq) + 3C6H6(l)
Assuming that Al(C6H5)3 was converted completely to products, what is the weight percent of Al(C6H5)3 in original 1.25-g sample?

Answer 1

This is a stoichiometry problem. Convert, using the coefficients in the balanced equation, 0.951 g C6H6 product to grams Al(C6H5)3. Then %Al(C6H5)3 = [g Al(C6H5)3/1.25)]*100 = ?

Answer 2

Is it 84.8%?

Answer 3

I got 83.8%. close enough. just make sure not to round your values until the very end (keep them in your calc).

Answer 4

I don't understand how you got to 83.8%??