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?

No.

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.

Yes, you can use the ideal gas law to solve this problem. The ideal gas law equation is:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)

To solve for the volume of carbon dioxide produced, you need to rearrange the ideal gas law equation to isolate V (volume). The rearranged equation becomes:

V = (nRT) / P

Now, let's plug in the given values into the equation:
- n = 0.979 mol (number of moles of calcium carbonate)
- R = 0.0821 L·atm/(mol·K)
- T = 0°C + 273.15 = 273.15 K (temperature in Kelvin)
- P = 1 atm (pressure)

V = (0.979 mol * 0.0821 L·atm/(mol·K) * 273.15 K) / 1 atm

By performing the calculation, you will find the volume of carbon dioxide produced in liters.