The idea here is to convert 60.0.0 g of Cl2 to moles of Cl2, to moles PCl5 using dimensional analysis and equation coefficients. We can do that as follows: 1 mol Cl2 = 70.906 g Cl2 10 mol Cl2 = 4 mol PCl5 So what you do is the following: [(60.0 g Cl2 )/1][(1 mol Cl2)/(70.906 g Cl2)][(4 mol PCl5)/(10 mol Cl2)] = 0.3384762 mol PCl5 or 0.338 mol PCl5 That meanns that 0.338 mol of PCl5 can be produced from 60.0 g of Cl2 and excess P4.
