Question

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

Answer 1

`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).