posted by Mohamed on .
What volume of carbon dioxide is produced when 0.979 mol of calcium carbonate reacts completely according to the following reaction at 0oC and 1 atm?
calcium carbonate (s)---> calcium oxide (s) + carbon dioxide (g)
Should I use the ideal gas law?
Yes but first use the regular stoichiometry process to determine mols CO2 produced.
is 10.97 L correct?
No. You answer is too low by exactly a factor of 2.
Don't you have to divide moles of calcium carbonate by 2 to get moles of carbon dioxide? and then multiply by 0.0821 and 273 and then divide by 1?
CaCO3 = CaO + CO2
mols CaCO3 = 0.979 from the problem.
Convert mols CaCO3 to mols CO2 using the coefficients in the balanced equation. That is 0.979 mols CaCO3 x (1 mol CO2/1 mol CaCO3) = 0.979 x 1/1 - 0.979 mols CO2.
Then you may use PV = nRT and solve for V OR you can remember that at STP (which is that listed in the problem of 1 atm and 0 C) 1 mole of a gas occupies 22.4L.
0.979 mol x 22.4 L/mol = 21.93 which rounds to 21.9 to three significant figures. L.
Or from PV = nRT
V = nRT/P = 0.979*0.08206*273/1 atm =21.93 which rounds to 21.9 L to 3 s.f.