Question

Elemental phosphorus can be prepared from calcium phosphate via the overall reaction

$\mathbf{2}\mathit{C}{\mathit{a}}_{\mathbf{3}}\mathbf{(}\mathbf{P}{\mathbf{O}}_{\mathbf{4}}{\mathbf{)}}_{\mathbf{2}}\mathbf{+}\mathbf{6}\mathit{S}\mathit{i}{\mathit{O}}_{\mathbf{2}}\mathbf{+}\mathbf{10}\mathit{C}\mathbf{\to }\mathbf{6}\mathit{C}\mathit{a}\mathit{S}\mathit{i}{\mathit{O}}_{\mathbf{3}}\mathbf{+}{\mathit{P}}_{\mathbf{4}}\mathbf{+}\mathbf{10}\mathit{C}\mathit{O}$

Calculate the minimum mass of$\mathit{C}{\mathit{a}}_{\mathbf{3}}\mathbf{(}\mathbf{P}{\mathbf{O}}_{\mathbf{4}}{\mathbf{)}}_{\mathbf{2}}$required to produced $\mathbf{69}\mathbf{.}\mathbf{8}\mathit{g}$${\mathit{P}}_{\mathbf{4}}\mathbf{.}$What mass of$\mathit{C}\mathit{a}\mathit{S}\mathit{i}{\mathit{O}}_{\mathbf{3}}$is generate as a byproduct ?

Short answer

$350g$of calcium phosphate and $393g$of calcium silicate.

Step-by-step solution

Formation of elemental phosphorus

Calcium phosphate, silicon oxide, and carbon react and form calcium silicate, phosphorus, and carbon monoxide.

One has to state the direct proportion between the “reactant and the product masses” in order to calculate the result needed

Using the unitary method

According to the balanced chemical reaction, the following relations are true:

$2molC{a}_3(P{O}_4{)}_2\to 1mol{P}_4$$\&6molCaSi{O}_3$

$2\times 310.2g\to 123.9g$$\&6\times 116.2g$

$\otimes g\to 69.8g$$\&yg$

The unknown masses of calcium phosphate as the reactant is needed and the by-product, calcium silicatecan be calculated using the direct proportion as stated above

 Step 3: Calculatethe mass of the product

• Calcium phosphate

The direct proportion can be solved for the mass of the reactant as an equation:

$$\begin{gathered} mC{a}_3(P{O}_4{)}_2=\frac{69.8}{123.9}\times 620.4 \\ =350g \end{gathered}$$

• Calcium silicate

The direct proportion can be solved for the mass of the byproduct as an equation:

$$\begin{gathered} mCaSi{O}_3=\frac{69.8}{123.9}\times 697.2 \\ =393g \end{gathered}$$

Therefore, $350g$of calcium phosphate required and $393g$of calcium silicate is generated.