How many moles of pcl5 can be produced from 60.0 g of cl2 (and excess p4)?

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.

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