Question

Phosphorus can be prepared from calcium phosphate by the following reaction:

$$\begin{gathered} {\mathbf{\text{2Ca}}}_{\mathbf{3}}{\mathbf{\left(}{\mathbf{\text{PO}}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\left(}\mathbf{\text{s}}\mathbf{\right)}{\mathbf{\text{+6SiO}}}_{\mathbf{2}}\mathbf{\left(}\mathbf{\text{s}}\mathbf{\right)}\mathbf{\text{+10C}}\mathbf{\left(}\mathbf{\text{s}}\mathbf{\right)}{\mathbf{\text{→6CaSiO}}}_{\mathbf{3}}\mathbf{\left(}\mathbf{\text{s}}\mathbf{\right)}{\mathbf{\text{+P}}}_{\mathbf{4}}\mathbf{\left(}\mathbf{\text{s}}\mathbf{\right)}\mathbf{\text{+}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{10CO}}\mathbf{\left(}\mathbf{\text{g}}\mathbf{\right)} \end{gathered}$$

Phosphorite is a mineral that contains${\mathbf{\text{Ca}}}_{\mathbf{3}}{\left({\text{PO}}_4\right)}_{\mathbf{2}}$ plus other non-phosphorus-containing compounds. What is the maximum amount of${\mathbf{\text{P}}}_{\mathbf{4}}$ ,that can be produced$\mathbf{1}\mathbf{\text{kg}}$ from of phosphorite ,if the phosphorite sample is $\mathbf{75}\mathbf{\%}{\mathbf{\text{Ca}}}_{\mathbf{3}}{\left({\text{PO}}_4\right)}_{\mathbf{2}}$by mass? Assume an excess of the other reactants.

Short answer

The maximum amount of ${\text{P}}_4$, that can be produced from$10\text{kg}$ of phosphorite if the phosphorite sample is $75\%{\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$by mass is$\mathbf{149}\mathbf{.}\mathbf{48}\mathbf{}\mathbf{\text{g}}$ .

Step-by-step solution

Definition

Themassof any material expressed in atomic mass unitsisquantitatively equivalent to its molar mass in grams per mole.

Calculate the mass ofCa(PO4)2  in mineral

Mineral containsof $\mathbf{75}\mathbf{\%}$mass,$$\begin{gathered} {\text{MassofCa}}_3{\left({\text{PO}}_4\right)}_2\text{=75\%Massofmineral} \\ =\frac{75}{100}1\times {10}^4\text{kg} \end{gathered}$$

$=\text{750kg}$

Balanced Equation

The balanced equation is

${\text{2Ca}}_3{\left({\text{PO}}_4\right)}_2\left(\text{s}\right){\text{+6SiO}}_2\left(\text{s}\right)\text{+10C}\left(\text{s}\right){\text{→6CaSiO}}_3\left(\text{s}\right){\text{+P}}_4\left(\text{s}\right)\text{+10CO}\left(\text{g}\right)$

Let’s convert $\mathbf{\text{Ca}}{\left({\text{PO}}_4\right)}_{\mathbf{2}}$to the number of moles:

$\text{n}\left(\text{Ca}{\left({\text{PO}}_4\right)}_2\right)\text{=750g}\left(\text{Ca}{\left({\text{PO}}_4\right)}_2\right)\text{×}\frac{\text{1mol}}{\text{310.172g}\left({\text{Ca}}_3{\left({\text{PO}}_4\right)}_2\right)}$

$=310.172\frac{\text{g}}{\text{mol}}$

$\text{m}\left({\text{P}}_4\right){\text{=1.209molP}}_4\text{×}\frac{123.64\text{g}}{1{\text{molP}}_4}$

Thus for the balance reactions,

Mol ratio is$\frac{{\text{1molP}}_4}{2{\text{molCa}}_3{\left({\text{PO}}_4\right)}_2}$

$\text{n}\left({\text{P}}_4\right){\text{=2.14molCa}}_3{\left({\text{PO}}_4\right)}_2\text{×}\frac{{\text{1molP}}_4}{2{\text{molCa}}_3{\left({\text{PO}}_4\right)}_2}$

$=1.209\text{mol}$

Determination of molar mass

Now, the molar mass:

$\left({\text{P}}_4\right)=4\times 30.91$

$=123.64\frac{\text{g}}{\text{mol}}$

$=149.48{\text{gP}}_4$

Therefore, the maximum amount of ${\text{P}}_4$is produced as $\mathbf{149}\mathbf{.}\mathbf{48}\mathbf{}\mathbf{\text{g.}}$