How Do i solve this chem problem?

Question

0.250 mol of methanol CH3OH(g) is completely burned according to the equation:

2CH3OH(l) + 3O2(g) ----> 2CO2(g) + 4H2O(g) + heat

What volume of CO2(g) measured at 298 K and 101 kPa is produced

Answer 1

PV=nRT

0.250mol CH3OH x 2mol CO2/2mol CH3OH = 0.250 mols (n)

298K (T)

101kPa x 1 atm/101.325kPa = 0.997 atm (P)

0.082057  (R)

Plug into equation to solve for volume(V):

0.997(V) = 0.250(0.082057)(298)

V=  [(0.250)(0.082057)(298)] / 0.997

V=6.132 L