Answer to Question #69835 in General Chemistry for sharmaake

Question

how many moles are in 87.6 g of aluminium sulphate, A1(SO4)3(K)

Answer 1

M(Al2(SO4)3)= 342.15 g/mol
n=m/M=(87.6 g)/(342.15 g/mol)=0.256 mol≅0.3 mol
Answer
In the anhydrous aluminum sulfate must be 0.3 moles.