Question

Carbon monoxide gas reacts with hydrogen gas to form methanol.
CO(g)+2H2(g)?CH3OH(g)
A 1.75L reaction vessel, initially at 305 K, contains carbon monoxide gas at a partial pressure of 232 mmHg and hydrogen gas at a partial pressure of 372mmHg . Identify the limiting reactant and determine the theoretical yeild of methonal in grams. Thank you!

Answer 1

The partial pressure of hydrogen is 372 mm Hg, which is 1.603 times the partial pressure of CO (232 mm Hg). This means that you have about 1.603 moles of hydrogen for each mole of CO.

If you look at the equation that you provided, you will see that you actually need 2 moles of hydrogen for each mole of CO. This means that you don't have enough hydrogen, which means hydrogen is the limiting reactant.

You have 372 mm Hg of hydrogen. The ideal gas law is PV = nRT.
P (of hydrogen only) = 232 mm Hg = 0.489atm (atm and mm Hg are both units of pressure, with 1 atm = 760 mm Hg)
V=1.75L
R=0.08206 (L*atm)/(mol*K)
T=305K

PV=nRT becomes 0.489*1.75=n*0.08206*305
So n = 0.0342 moles of hydrogen. Divide that by 2 to get moles of methanol, then use the molar mass to find the number of grams of methanol.

(I didn't actually mean that you had 1.603 moles of hydrogen overall. I just meant that that was how many moles of hydrogen you had for each 1 mole of CO.)