How many moles of pcl5 can be produced from 28.0 g of p4 (and excess cl2)?

P₄ + 10Cl₂ ---> 4PCl₅
stoichiometry of P₄ to PCl₅ is 1:4
number of moles of P₄ reacted - 28.0 g / 124 g/mol = 0.22 mol 
Cl₂ is in excess therefore P₄ is the limiting reactant, amount of product formed depends on amount of limiting reactant present 
according to molar ratio of 1:4
number of PCl₅ moles formed  -0.22 mol x 4 = 0.88 mol 
0.88 mol of PCl₅ is formed 

Ithink the answer is d