If you start with 17.2 g of Al2(SO3)3 and you produce 6.83 gAl(OH)3, what is your percent yield for this reaction?

Al2(SO3)3 + 6 NaOH → 3 Na2SO3 + 2 Al(OH)3

Explanation:

Hello,

In this case, since the aluminum sulfite, whose molar mass is 294.153 g/mol, is in a 1:2 molar ratio with aluminum hydroxide, whose molar mass is 78 g/mol, we can perform the following stoichiometric procedure to compute the theoretical yield of aluminum hydroxide to furtherly compute the percent yield:

So we compute the percent yield considering the actual yield is 6.83 g:

Regards.

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