Question

Carbon monoxide gas reacts with hydrogen gas to form methanol.
CO(g)+2H2(g)→CH3OH(g)
A 1.05 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 357 mmHg .

Identify the limiting reactant and determine the theoretical yield of methanol in grams

Answer 1

Solution :-

CO(g)+2H2(g) ------ > CH3OH(g)

Volume = 1.05 L

Temperature = 305 K

Pressure = (232 + 357 mmHg )* 1 atm /760 mmHg = 0.775 atm

Lets first calculate the total moles of the reactants in the vessel

PV= nRT

PV/RT = n

0.775 atm * 1.05 L / 0.08206 L atm per mol K * 305 K = n

0.03251 mol = n

Now lets calculate the moles of each gas using the partial pressures

Total pressure = 232 mmHg + 357 mmHg = 589 mmHg

Moles of CO = (partial pressure / total pressure ) *total moles

                         = (232 mmHg / 589 mmHg) * 0.03251 mol

                         = 0.012805 mol CO

Moles of H2 = total moles - moles of CO

                       = 0.019705 mol H2

Now lets calculate the moles of the CH3OH that can be oroduced from each reactant using the mole ratio of the each reactant

(0.012805 mol CO * 1 mol CH3OH / 1 mol CO) = 0.012805 mol CH3OH

(0.019705 mol H2 * 1 mol CH3OH / 2 mol H2)= 0.009825 mol CH3OH

H2 gives less moles of product

Therefore H2 is the limiting reactant

So the moles of the CH3OH that can be formed are 0.009825 mol CH3OH

Now lets calculate the mass of CH3OH

Mass= moles * molar mass

         = 0.009825 mol CH3OH * 32.04 g per mol

         = 0.3157 g CH3OH

So the theoretical mass of CH3OH that can be formed = 0.3157 g