Question #6e423

1 Answer
Jan 22, 2017

Answer:

66.2 % (m/m) of the #"CaCO"_3# will react.

Explanation:

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 %#