Question

1. How many moles of PCl5 can be produced from 30.0 g of P4 (and excess Cl2)? 2.How many moles of PCl5 can be produced from 59.0 g of Cl2 (and excess P4)? 3.What mass of PCl5 will be produced from the given masses of both reactants?

Answer 1

1.
Molar mass of P4= 124 g/mol
mass of P4 = 30 g
number of moles of P4 =mass / molar mass
                    = 30/124
                     =0.242 mol
P4+ 10CL2 ---> 4PCl5
1 mol of P4 will form 4 mol of PCL5
So,
number of moles of PCl5 = 4*0.242 g =0.97 mol

2.
Molar mass of P4= 124 g/mol
mass of P4 = 59 g
number of moles of P4 =mass / molar mass
                    = 59/124
                     =0.476 mol
P4+ 10CL2 ---> 4PCl5
1 mol of P4 will form 4 mol of PCL5
So,
number of moles of PCl5 = 4*0.476 g =1.9 mol

3.
Molar mass of PCl5 = 208 g/mol
For 1:
number of moles of PCl5 = 0.97 mol
mass of PCl5 = 0.97*208= 202 g

For 2:
number of moles of PCl5 =1.9 mol
mass of PCl5 = 1.9*208= 395 g