The answers are V_(H_2) = 43.1 L and V_(CO) = 21.5 L.
Starting from the ideal gas law PV = nRT and from the chemical reaction's balanced equation
CO(g) + 2H_2(g) -> CH_3OH(g), one can see that we get a 1:1 mole ratio for CO and CH_3OH, and a 2:1 mole ratio forH_2 and CH_3OH.
The number of moles of CH_3OH can be calculated from
n_(CH_3OH) = m_(CH_3OH)/(32 g/(mol)) =(25.8 g)/(32 g/(mol)) = 0.81 moles.
Subsequently, the number of moles for CO and H_2 are
n_(CO) = 0.81 moles, and n_(H_2) = 1.62 moles.
Therefore, the volume of H_2 required for the reaction to take place is
V_(H_2) = (n_(H_2)RT)/P = (1.62 * 0.082 * 319.15)/ (748/760) = 43.1 L ( I've converted pressure into atm anddegrees Celsius into K).
The volume of CO required will be
V_(CO) = (n_(CO) RT)/P = (0.81 * 0.082 * 319.15)/(748/760) = 21.5 L, which is approximately half of the volume of H_2 determined (half the number of moles -> half the volume);
