Question

Carbon monoxide gas reacts with hydrogen gas to form methanol. CO(g)+2H2(g)→CH3OH(g) A 1.10 L 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 396 mmHg .Identify the limiting reactant and determine the theoretical yield of methanol in grams.

Answer 1

CO(g) + 2H₂(g) -------> CH₃OH(g)

mole fraction of CO = pressure of CO/total pressure of gas mixture = 232/(232+396) = 0.369

mole fraction of H₂ = pressure of H₂/total pressure of gas mixture = 396/(232+396) = 0.631

Now, total pressure of the gas mixture = 232 + 396 = 628 mm Hg = 0.826 atm

Applying Ideal Gas Equation , we get

Total moles of gas mixture, n = (P*V)/R*T) = (0.826*1.1)/(0.0821*305) = 0.0363

Thus, moles of CO = mole fraction*total number of moles = 0.0134

moles of H₂ = mole fraction*total number of moles = 0.0229

Now, as per the balanced reaction, CO & H₂ reacts in the molar ratio of 1:2

Thus, moles of H₂ required for 0.0134 moles of CO = 0.0268

Clearly, H₂ is the limiting reagent

Thus, moles of methanol formed = (1/2)*moles of H₂ reacting = 0.01145

Now, molar mass of methanol = 32 g/mole

Thus, mass of methanol formed = moles*molar mass = 0.3665 g