_A2(SO3)3+_NaOH->_Na2SO3+_Al(OH)3

Question

if you start with 389.4 g of Al2(SO3)3 and you produce 212.4 g of Na2SO3, what is your percent yield for this reaction?

Answer 1

First thing you always do; balance the equation.
Convert 389.4 g Al2(SO4)3 to mols. mol = grams/molar mass.
User the coefficients in the balanced equation to convert mols Al2(SO4)3 to mols Na2SO3.
Convert mols Na2SO3 to grams. g = mols x molar mass. This is the theoretical yield (TY). The actual yield(AY) listed in the problem is 212.4.
%yield = (AY/TY)*100 = ?