Question

Phosphoric acid is made industrially by the reaction of fluorapatite,${\mathbf{Ca}}_{\mathbf{5}}{\left({\mathrm{PO}}_4\right)}_{\mathbf{3}}\mathbf{F}$, in phosphate rock with sulphuric acid: ${\mathbf{Ca}}_{\mathbf{5}}{\left({\mathrm{PO}}_4\right)}_{\mathbf{3}}\mathbf{}\mathbf{F}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{5}{\mathbf{H}}_{\mathbf{2}}{\mathbf{SO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{10}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{\ell }\mathbf{\right)}\mathbf{\to }\mathbf{3}{\mathbf{H}}_{\mathbf{3}}{\mathbf{PO}}_{\mathbf{4}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{5}\left({\mathrm{CaSO}}_4·2H_{2}O\right)\mathbf{\left(}\mathbf{s}\right)+\mathrm{HF}\left(aq\right)$.

What volume of 6.3Mphosphoric acid is generated by the reaction of 2.2metric tons (2200kg)of fluorapatite?

Short answer

The volume of 6.3M phosphoric acid generated by the reaction of 2.2 metric tons (2200kg) of fluorapatite is 2100L.

Step-by-step solution

The concept of the number of moles

A mole is defined as the mass of the substance which consists of an equal quantity of basic units.One mole of any substance is equal to the value of$\mathbf{6}\mathbf{.}\mathbf{023}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}$(Avogadro number). It can be used to measure the products obtained from the chemical reaction. The unit is denoted by mol.

The formula for the number of moles formula is expressed as:

$\mathbf{n}\mathbf{=}\frac{\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{substance}}{\mathbf{Mass}\mathbf{}\mathbf{of}\mathbf{}\mathbf{one}\mathbf{}\mathbf{mole}}$

Find the number of moles of  Ca5(PO4)3F

Substitute the values of mass and molar mass of ${\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}$in the above formula as:

$\begin{array}{l}\mathrm{n}[\mathrm{C}{\mathrm{a}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}]=\frac{2200\mathrm{kg}}{504.31{\mathrm{gmol}}^{-1}}\times \frac{{10}^3\mathrm{g}}{1\mathrm{kg}}\\ \mathrm{n}[\mathrm{C}{\mathrm{a}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}]=4.4\times {10}^3\mathrm{mol}\end{array}$

Find the number of moles of phosphoric acid

The number of moles of phosphoric acid is calculated using stoichiometry as follows:

$\begin{array}{l}\mathrm{n}\left[{\mathrm{H}}_3{\mathrm{PO}}_4\right]=4.4\times {10}^3{\mathrm{molCa}}_{\mathrm{s}}{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}\times \frac{3\mathrm{mol}{\mathrm{H}}_3{\mathrm{PO}}_4}{1\mathrm{mol}{\mathrm{Ca}}_5{\left({\mathrm{PO}}_4\right)}_3\mathrm{F}}\\ \mathrm{n}\left[{\mathrm{H}}_3{\mathrm{PO}}_4\right]=1.3\times {10}^4\mathrm{mol}\end{array}$

Determine the volume of phosphoric acid required to react with  

Substitute the values to find the volume of phosphoric acid required:

$\begin{array}{l}\mathrm{V}\left[{\mathrm{H}}_3{\mathrm{PO}}_4\right]=1.3\times {10}^4{\mathrm{molH}}_3{\mathrm{PO}}_4\times \frac{\mathrm{L}}{6.3{\mathrm{molH}}_3{\mathrm{PO}}_4}\\ \mathrm{V}\left[{\mathrm{H}}_3{\mathrm{PO}}_4\right]=2.1\times {10}^3\mathrm{L}\\ \mathrm{V}\left[{\mathrm{H}}_3{\mathrm{PO}}_4\right]=2100\mathrm{L}\end{array}$

Thus, the volume of 6.3M phosphoric acid generated by the reaction of 2.2 metric tons (2200kg) of fluorapatite is 2100L.