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 for#H_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);
