What's the mass of Mg3(PO4)2 in 0,4 mol?

Question

What's the mass of Mg3(PO4)2 in 0,4 mol?

Answer 1

M (Mg3(PO4)2) = (24*3)+((31+16*4)*2)) = 262 г/моль

m (Mg3(PO4)2) = 262 г/моль * 0,4 моль = 104,8 г