Answer
mass percentage of $CaCO_{3}$ = 83.7 %
Work Step by Step
The strategy: 1- find the mass of calcium carbonate that produce 496 mg of calcium oxalate
2- Find mass percentage of calcium carbonate in the limestone.
Ca C$O_{3}$ (s) + $H_{2}C_{2}O_{4}$ (aq) ${\longrightarrow} $ Ca$C_{2}O_{4}$ (s) + $H_{2}O$ (l) + C$O_{2}$ (g)
Step 1: to find the mass of Ca C$O_{3}$ when we know the amount 469 mg of Ca$C_{2}O_{4}$ :
g Ca$C_{2}O_{4}$ ${\rightarrow}$ mol Ca$C_{2}O_{4}$ ${\rightarrow}$ mol Ca C$O_{3}$${\rightarrow}$ g Ca C$O_{3}$
The conversion factors are:
$\frac{1 mol (CaC_{2}O_{4})}{Molar mass (CaC_{2}O_{4})}$ to convert to moles Ca$C_{2}O_{4}$, where Molar mass Ca$C_{2}O_{4}$:
Molar mass Ca$C_{2}O_{4}$= 1 x 40 + 2 x 12 + 4 x 16 = 128 g
$\frac{1 mol Ca CO_{3})}{1 mol (CaC_{2}O_{4})}$, to convert the moles of Ca$C_{2}O_{4}$ to mol of Ca C$O_{3}$.
$\frac{Molar mass(Ca CO_{3})}{1 molCa CO_{3})}$ to convert to g of Ca C$O_{3}$, where the molar mass of Ca C$O_{3}$ is:
Molar mass Ca C$O_{3}$ = 1 x 40 + 1 x 12 + 3 x 16 = 100 g
Now we use the conversion factors to find the mas of Ca C$O_{3}$:
496 x $10^{-3}$ g Ca$C_{2}O_{4}$ x $\frac{1 mol (CaC_{2}O_{4})}{Molar mass (CaC_{2}O_{4})}$ x$\frac{1 mol Ca CO_{3})}{1 mol (CaC_{2}O_{4})}$ x $\frac{Molar mass(Ca CO_{3})}{1 molCa CO_{3})}$=
469 x $10^{-3}$ g Ca$C_{2}O_{4}$ x $\frac{1 mol (CaC_{2}O_{4})}{128 (CaC_{2}O_{4})}$ x$\frac{1 mol Ca CO_{3})}{1 mol (CaC_{2}O_{4})}$ x $\frac{100 g(Ca CO_{3})}{1 molCa CO_{3})}$= 366.4 $10^{-3}$ g $CaCO_{3}$ or 366.4 mg $CaCO_{3}$.
Step 2 - Find the mass percentage of $CaCO_{3}$ in limestone.
mass percentage of $CaCO_{3}$ = $\frac{mass _(CaCO_{3})}{mass_(limestone)}$ x 100%
mass percentage of $CaCO_{3}$ = $\frac{366.4g}{438g}$ x 100%
mass percentage of $CaCO_{3}$ = 83.7 %
