Answer
150 g $Ca_{3}(PO_{4})_2$.
Work Step by Step
Balanced chemical reaction:
2 $Ca_{3}(PO_{4})_2$ (s) + 6 Si$O_{2}$(s) + 10 C(s) ${\longrightarrow}$ $P_{4}$ (g) + 6 $CaSiO_{3}$ (l) + 10 CO (g)
The strategy to find the mass of $Ca_{3}(PO_{4})_2$ required to produce 30.0 g $P_{4}$ is:
g $P_{4}$ ${\rightarrow}$ mol $P_{4}$ ${\rightarrow}$ mol $Ca_{3}(PO_{4})_2$ ${\rightarrow}$ g$Ca_{3}(PO_{4})_2$
The conversion factors are:
$\frac{1 mol (P_{4})}{Molar mass (P_{4})}$ to convert to moles $P_{4}$, where Molar mass $P_{4}$:
Molar mass $P_{4}$ = 4 x 31.0 g = 124 g
$\frac{2 mol (Ca_{3}(PO_{4})_2)}{1 mol (P_{4})}$, to convert the moles of $P_{4}$ to mol of $Ca_{3}(PO_{4})_2$.
$\frac{Molar mass(Ca_{3}(PO_{4})_2)}{1 mol(Ca_{3}(PO_{4})_2)}$ to convert to g of $Ca_{3}(PO_{4})_2$, where the molar mass of $Ca_{3}(PO_{4})_2$ is:
molar mass of $Ca_{3}(PO_{4})_2$ = 3 x 40 + 2 x 31.0 + 8 x 16 = 310 g
Now we use of the conversion factors to find the mass of $Ca_{3}(PO_{4})_2$:
30.0 g $P_{4}$ x $\frac{1 mol (P_{4})}{Molar mass (P_{4})}$ x $\frac{2 mol (Ca_{3}(PO_{4})_2)}{1 mol (P_{4})}$ x $\frac{Molar mass(Ca_{3}(PO_{4})_2)}{1 mol(Ca_{3}(PO_{4})_2)}$
30.0 g $P_{4}$ x $\frac{1 mol (P_{4})}{124 g (P_{4})}$ x $\frac{2 mol (Ca_{3}(PO_{4})_2)}{1 mol (P_{4})}$ x $\frac{310 g(Ca_{3}(PO_{4})_2)}{1 mol(Ca_{3}(PO_{4})_2)}$ = 150 g $Ca_{3}(PO_{4})_2$
