Step 1. Calculate P_"CO₂" at equilibrium
Start by writing the chemical equation and preparing an ICE table.
color(white)(mmmm)"CaCO"_3"(s)" ⇌ "CaO(s)" + "CO"_2"(g)"; K_text(P) = 1.16
"I/atm:"color(white)(mmmmmmmmmmmmml)0
"C/atm:"color(white)(mmmmmmmmmmmmll)"+"x
"E/atm:"color(white)(mmmmmmmmmmmmm)x
K_"P" = x = 1.16
∴ P_"CO₂" = xcolor(white)(l) "atm" = "1.16 atm"
2. Use the Ideal Gas Law to calculate the moles of "CO"_2.
PV = nRT
n = (PV)/(RT) = (1.16 color(red)(cancel(color(black)("atm"))) × 10.1 color(red)(cancel(color(black)("L"))))/("0.082 06" color(red)(cancel(color(black)("L·atm·K"^"-1")))"mol"^"-1" × 1073.15 color(red)(cancel(color(black)("K")))) = "0.1330 mol"
Step 3. Calculate the moles of "CaCO"_3 reacted.
"Moles of CaCO"_3 = 0.1330 color(red)(cancel(color(black)("mol CO"_2))) × ("1 mol CaCO"_3)/(1 color(red)(cancel(color(black)("mol CO"_2)))) = "0.1330 mol CaCO"_3
Step 4. Calculate the mass of "CaCO"_3 reacted.
"Mass of CaCO"_3 = 0.1330 color(red)(cancel(color(black)("mol CaCO"_3))) × ("100.09 g CaCO"_3)/(1 color(red)(cancel(color(black)("mol CaCO"_3)))) = "13.31 g CaCO"_3
Step 5. Calculate the percent by mass of "CaCO"_3
"% by mass" = (13.31 color(red)(cancel(color(black)("g CaCO"_3))))/(20.1 color(red)(cancel(color(black)("g CaCO"_3)))) × 100 % = 66.2 %
