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

Jun 12, 2017

$0.310$ ${\text{mol PCl}}_{5}$

#### Explanation:

We're asked to calculate the number of moles of ${\text{PCl}}_{5}$ formed from excess ${\text{P}}_{4}$ and $55.0$ ${\text{g CL}}_{2}$.

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

${\text{P"_4 (s) + 10"Cl"_2 (g) rarr 4"PCl}}_{5} \left(s\right)$

Using the molar mass of ${\text{Cl}}_{2}$, let's convert from grams to moles, and then use the coefficients in the equation to convert to moles of ${\text{PCl}}_{5}$:

$55.0$ 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 ${\text{PCl}}_{5}$ will form (assuming the reaction goes to completion).