Question

Phosphorous reacts with chlorine gas to produce phosphorous pentachloride. Calculate the mass of product produced when 25.0 g of phosphorous reacts with 25.0 grams of chlorine. Calculate the mass of product produced if the reaction occurred with a 70.5 percent yield.

Answer 1

Answer : The mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.

Explanation : Given,

Mass of P = 25 g

Mass of $Cl_2$ = 25 g

Molar mass of P = 30.97 g/mole

Molar mass of $Cl_2$ = 71 g/mole

Molar mass of $PCl_5$ = 208.24 g/mole

First we have to calculate the moles of $P$ and $Cl_2$.

$\text{Moles of }P=\frac{\text{Mass of }P}{\text{Molar mass of }P}=\frac{25g}{30.97g/mole}=0.807moles$

$\text{Moles of }Cl_2=\frac{\text{Mass of }Cl_2}{\text{Molar mass of }Cl_2}=\frac{25g}{71g/mole}=0.352moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2P+5Cl_2\rightarrow 2PCl_5$

From the balanced reaction we conclude that

As, 5 moles of $Cl_2$ react with 2 moles of $P$

So, 0.352 moles of $Cl_2$ react with $\frac{2}{5}\times 0.352=0.1408$ moles of $P$

That means, in the given balanced reaction, $Cl_2$ is a limiting reagent and it limits the formation of products and $P$ is an excess reagent because the given moles are more than the required moles.

Now we have to calculate the moles of $PCl_5$.

As, 5 moles of $Cl_2$ react with 2 moles of $PCl_5$

So, 0.352 moles of $Cl_2$ react with $\frac{2}{5}\times 0.352=0.1408$ moles of $PCl_5$

Now we have to calculate the mass of $PCl_5$.

$\text{Mass of }PCl_5=\text{Moles of }PCl_5\times \text{Molar mass of }PCl_5$

$\text{Mass of }PCl_5=(0.1408mole)\times (208.24g/mole)=29.32g$

Now we have to calculate the mass of product produced (actual yield).

$\%\text{ yield of }PCl_5=\frac{\text{Actual yield of }PCl_5}{\text{Theoretical yield of }PCl_5}\times 100$

$70.5=\frac{\text{Actual yield of }PCl_5}{29.32g}\times 100$

$\text{Actual yield of }PCl_5=20.67g$

Therefore, the mass of product produced if the reaction occurred with a 70.5 percent yield will be, 20.67 grams.