Question

Consider the following unbalanced reaction:

${\mathbf{\text{P}}}_{\mathbf{4}}\left(s\right)\mathbf{+}{\mathbf{\text{F}}}_{\mathbf{2}}\left(g\right)\mathbf{\to }{\mathbf{\text{PF}}}_{\mathbf{3}}\left(g\right)$
How many grams of are needed to produce 120.g of ${\mathbf{\text{PF}}}_{\mathbf{3}}$ if the reaction has a 78.1% yield?

Short answer

$\mathbf{99}\mathbf{.}\mathbf{522}\mathbf{}\mathbf{\text{g}}$ of ${\text{F}}_2$ are needed to produce 120g of ${\text{PF}}_3$ if the reaction has a 78.1% yield.

Step-by-step solution

Definition

A balanced chemical equation is one in whichthe number of atoms present on the reactant side of the equation equals the number of atoms present on the product side of the equation.

Determining of number of moles

Here the molar mass ${\text{PF}}_3$$=30.973+3\left(18.998\right)$

=$87.967\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

$\mathbf{\text{Numberofmoles=}}\frac{\mathbf{\text{mass}}}{\mathbf{\text{molarmass}}}$

$=\frac{120}{87.967}$

$=1.364\mathrm{mol}$

Determining balanced equation

${\mathrm{P}}_4\left(\mathrm{s}\right)+{\mathrm{F}}_2\left(\mathrm{g}\right)\to {\mathrm{PF}}_3\left(\mathrm{g}\right)$

Here, for the production of 4 moles of${\text{PF}}_3$ , 6 moles of ${\text{F}}_2$are needed.

Hence, theyield is only 78.1%.

$\left(78.1\%\right)\times \left(x\right)=120\text{g}$

role="math" localid="1648641070492" $x=\frac{120}{78.1}\times 100$

$=153.6491\text{g}$

Determining grams of F2  needed

Number moles production of${\text{PF}}_3$, $=\frac{153.6491g}{87.967}$

$=1.746\text{mol}$

${\text{PF}}_3$ Production will be, $=\frac{1.746}{4}\times 6$

$=2.619\text{Molesof}{\mathrm{F}}_2$

1 mole${\text{F}}_2=38.00\text{g}$

${\text{F}}_2=\left(2.619\right)\times \left(38\right)$

$=99.522\text{g}$

Therefore,role="math" localid="1648640671949" $\mathbf{99}\mathbf{.}\mathbf{522}\mathbf{\text{g}}$of${\text{F}}_2$are needed.