13. To obtain phosphorus, calcium phosphate was previously treated with sulfuric acid, the resulting orthophosphoric acid was mixed with carbon and calcined. In this process, orthophosphoric acid turned into metaphosphoric acid, which, upon interaction with carbon, yielded phosphorus, hydrogen, and carbon monoxide. Illustrate all stages of phosphorus production by this method in the form of reaction equations. Calculate the mass of the water-soluble salt that forms when 31 g of calcium phosphate reacts with 17.75 ml of a $69\%$ sulfuric acid solution (density of the solution 1.60 g/ml). (10 points)
13. $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}+3 \mathrm{H}_{2} \mathrm{SO}_{4}=2 \mathrm{H}_{3} \mathrm{PO}_{4}+3 \mathrm{CaSO}_{4} \cdot \mathrm{H}_{3} \mathrm{PO}_{4}=\mathrm{HPO}_{3}+\mathrm{H}_{2} \mathrm{O}$ (heating). $2 \mathrm{HPO}_{3}+6 \mathrm{C}=2 \mathrm{P}+\mathrm{H}_{2}+6 \mathrm{CO}$. Let's find the mass and number of moles of $\mathrm{H}_{2} \mathrm{SO}_{4}$ $17.75 * 1.60 * 0.69=19.6 \mathrm{r} ; \mathrm{n}\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)=19.6 / 98=0.2$ moles. Let's find the number of moles of $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2} \mathrm{n}\left(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right)=31 / 310=0.1$ moles. Write the equation according to the reagent ratio, taking into account that the water-soluble salt is $\mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}-$ calcium dihydrogen phosphate $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}+2 \mathrm{H}_{2} \mathrm{SO}_{4}=\mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}+$ $2 \mathrm{CaSO}_{4}$. Let's find the mass of $\mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2} \mathrm{~m}\left(\mathrm{Ca}\left(\mathrm{H}_{2} \mathrm{PO}_{4}\right)_{2}\right)=0.1 * 234=23.4$ g.