How many moles of pcl5 can be produced from 55.0 g of cl2 (and excess p4)? express your answer to three significant figures and include the appropriate units?

The balanced equation for the reaction is,
P₄ + 10Cl₂ ---> 4PCl₅
Stoichiometry of Cl₂ to PCl₅ 10:4
The number of Cl₂ moles reacted - 55.0 g/ 71 g/mol = 0.77 mol
Cl₂ is the limiting reactant, amount of product formed depends on amount of limiting reactant present.
10 mol of Cl₂ produces - 4 mol of PCl₅
when 0.77 mol of Cl₂ reacts - 4/10 x 0.77 = 0.308 mol 
Therefore 0.308 mol of PCl₅ is formed 

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The answer is nitrogen-containing bases