# Question 6e423

Jan 22, 2017

66.2 % (m/m) of the ${\text{CaCO}}_{3}$ will react.

#### Explanation:

Step 1. Calculate ${P}_{\text{CO₂}}$ 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
$\text{I/atm:} \textcolor{w h i t e}{m m m m m m m m m m m m m l} 0$
$\text{C/atm:"color(white)(mmmmmmmmmmmmll)"+} x$
$\text{E/atm:} \textcolor{w h i t e}{m m m m m m m m m m m m m} x$

${K}_{\text{P}} = x = 1.16$

${P}_{\text{CO₂" = xcolor(white)(l) "atm" = "1.16 atm}}$

2. Use the Ideal Gas Law to calculate the moles of ${\text{CO}}_{2}$.

$P V = n R T$

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 ${\text{CaCO}}_{3}$ reacted.

${\text{Moles of CaCO"_3 = 0.1330 color(red)(cancel(color(black)("mol CO"_2))) × ("1 mol CaCO"_3)/(1 color(red)(cancel(color(black)("mol CO"_2)))) = "0.1330 mol CaCO}}_{3}$

Step 4. Calculate the mass of ${\text{CaCO}}_{3}$ reacted.

${\text{Mass of CaCO"_3 = 0.1330 color(red)(cancel(color(black)("mol CaCO"_3))) × ("100.09 g CaCO"_3)/(1 color(red)(cancel(color(black)("mol CaCO"_3)))) = "13.31 g CaCO}}_{3}$

Step 5. Calculate the percent by mass of ${\text{CaCO}}_{3}$

"% 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 %#