Phosphorus occurs naturally in the form of fluorapatite, $\mathrm{CaF}_{2} \cdot 3 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}.$ The dot indicates 1 part $\mathrm{CaF}_{2}$ to 3 parts $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}.$ This mineral is reacted with an aqueous solution of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in the preparation of a fertilizer. The products are phosphoric acid, hydrogen fluoride, and gypsum, $\mathrm{CaSO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}.$ Write the balanced equation describing this process.

Question

Quizlet Admin · Accepted Answer

1. The equation to be balanced is:
\begin{equation*}
CaF_2 \cdot 3Ca_3(PO_4)_2 + H_2SO_4 \cdot xH_2Oightarrow H_3PO_4 + HF + CaSO_4 \cdot 2H_2O
\end{equation*}
2. Balance the Ca first:
\begin{equation*}
CaF_2 \cdot 3Ca_3(PO_4)_2 + H_2SO_4\cdot xH_2O ightarrow H_3PO_4 + HF + 10(CaSO_4 \cdot 2H_2O)
\end{equation*}
3. Balance the F:
\begin{equation*}
CaF_2 \cdot 3Ca_3(PO_4)_2 + H_2SO_4 \cdot xH_2O ightarrow 6H_3PO_4 + 2HF + 10(CaSO_4 \cdot 2H_2O)
\end{equation*}
4. Balance the phosphoric acid:
\begin{equation*}
CaF_2 \cdot 3Ca_3(PO_4)_2 + H_2SO_4 \cdot xH_2O ightarrow 6H_3PO_4 + 2HF + 10(CaSO_4 \cdot 2H_2O)
\end{equation*}
5. Balance the sulfuric acid:
\begin{equation*}
CaF_2 \cdot 3Ca_3(PO_4)_2 + 10(H_2SO_4 \cdot xH_2O) ightarrow 6H_3PO_4 + 2HF + 10(CaSO_4 \cdot 2H_2O)
\end{equation*}
6. Count all the non O and H atoms, and make sure it is equal in both sides: 
\begin{equation*}
    \mbox{10 Ca, 2 F, 6 P, 10 S}
\end{equation*}
7. Since we dont know how hydrated the sulfuric acid is, the coefficient is written as x. We can count all the O atoms in both sides and they must be equal:
\begin{align*}
    24 + 40 + 10x &= 24 + 40 + 20 \
    10x &= 20 \
    x &= 2
\end{align*}
The coefficient for H	extsubscript{2}O of sulfuric acid is 2.\\
8. Thus the balanced equation is:
\begin{equation*}
CaF_2 \cdot 3Ca_3(PO_4)_2 + 10(H_2SO_4 \cdot 2H_2O) ightarrow 6H_3PO_4 + 2HF + 10(CaSO_4 \cdot 2H_2O)
\end{equation*}

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