(50 ~g CaCO _ 3 ) is allowed to react with (49 ~g H _ 3 PO _ 4 )...

Given Reaction: (3 $\mathrm{CaCO}_{3}+2 \mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_2+\varepsilon) Molar Masses - (\mathrm{CaCO}_{3}$: 100 $\mathrm{~g}$/ $\mathrm{mol}$ - ($\mathrm{H}_{3} \mathrm{PO}_{4}$: 98 $\mathrm{~g}$/ $\mathrm{mol}$ Moles of Reactants - Moles of ($\mathrm{CaCO}_{3}) :$$\begin{aligned} & \frac{50 \mathrm{~g}}{100 \mathrm{~g} / \mathrm{mol}}=0.5 \mathrm{moles}\\ & \frac{49 \mathrm{~g}}{98 \mathrm{~g} / \mathrm{mol}}=0.5 \mathrm{moles} \end{aligned}$$) Limiting Reagent$ From $the stoichiometry: - Required: 3 moles of ($\mathrm{CaCO}_{3}$for 2 moles of ($\mathrm{H}_{3} $\mathrm{PO}_{4}$. - Since both reactants are present in a 1:1 ratio, neither is limiting for the given amounts. Products Formed - Moles of ($\mathrm{CaCO}_{3}) used$=0.5$moles$→$Produces 0.167 moles of (\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_2), 0.5 moles of (\mathrm{H}_{2}$\mathrm{O}$, and 0.5 moles of ($\mathrm{CO}_{2}). Masses - Salt formed: Mass of (\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_2=0.167 \times310 \mathrm{~g}/ $\mathrm{r}- Unreacted reagent: - 0.5 moles of$(\mathrm{H}_{3} \mathrm{PO}_{4}) reacts with 0.5 moles of ($\mathrm{CaCO}_{3}$ →) No unreacted reagent. - Incorrect interpretation earlier; 32.67 g unreacted is incorrect. (Incorrect: B)