Question

How many moles of PCl5 can be produced from 60.0 g of Cl2 (and excess P4)?
Express your answer to three significant figures and include the appropriate units.

Answer 1

2P + 5Cl2 => 2PCl5
m(Cl2 )= 60.0 g
m(PCl5) - ?
n(Cl2) = m(Cl2)/ M(Cl2) = 60.0 g / 71 g/mol = 0.845 mol
2n(PCl5) = 5n(Cl2 )
n(PCl5) = 2n(Cl2 )/ 5 = 2*0.845 mol/ 5 = 0.338 mol
m (PCl5) = n(PCl5) *M(PCl5) = 0.338 mol*208 g/mol = 70.304 g