How many moles of pcl5 can be produced from 53.0 g of cl2 (and excess p4)?

Calculate the number of moles of PCl5 that can be produced from 23.0 g of P4 (and excess).

P4 + 10 Cl2 → 4 PCl5
 
Molecular weight (MW) of p=30.97 g/mole


1 mol P4 = 123.8952 g P4
1 mol P4 = 4 mol PCl5

[(23.0 g P4)/1][(1 mol P4)/(123.8952 g P4)][(4 mol PCl5)/(1 mol P4)] = 0.7425 mol PCl5
 
Calculate the number of moles of PCl5 that can be produced from 53.0 g of Cl2 (and excess P4)

Molecular weight (MW) of Cl=70.906g/mole

1 mol Cl2 = 70.906 g Cl2
10 mol Cl2 = 4 mol PCl5

[(53.0 g Cl2)/1][(1 mol Cl2)/(70.906 g Cl2)][(4 mol PCl5)/(10 mol Cl2)] = 0.299 mol PCl5