How many moles of #PCl_5# can be produced from 55.0 g of #Cl_2# and excess #P_4#?

1 Answer
Jun 12, 2017

Answer:

#0.310# #"mol PCl"_5#

Explanation:

We're asked to calculate the number of moles of #"PCl"_5# formed from excess #"P"_4# and #55.0# #"g CL"_2#.

Let's start by writing the chemical equation for this reaction:

#"P"_4 (s) + 10"Cl"_2 (g) rarr 4"PCl"_5(s)#

Using the molar mass of #"Cl"_2#, let's convert from grams to moles, and then use the coefficients in the equation to convert to moles of #"PCl"_5#:

#55.0# #cancel("g Cl"_2)((1cancel("mol Cl"_2))/(70.90cancel("g Cl"_2)))((4"mol PCl"_5)/(10cancel("mol Cl"_2))) = color(red)(0.310# #color(red)("mol PCl"_5#

Thus, with excess phosphorus, #0.310# moles of #"PCl"_5# will form (assuming the reaction goes to completion).