Question

Carbon monoxide gas reacts with hydrogen gas to form methanol via the following reaction:

CO(g)+2H2(g)→CH3OH(g)

A 1.70 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 355 mmHg .

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

Answer 1

PV No of moles of the CO gas, n = (232 mmHg)(1.70 L) (62.4 LmmHg K-'m011)(305 K) = 0.0207 mol PV No of moles of the H, gas, n = 27 (355 mmHg) (1.70 L) (62.4 LmmHg K-'m01-1)(305 K) = 0.0317 mol - - - - - - - - - - - - - - - - - - - - - - - - 1 mol CH OH No of moles of CH,OH produced from CO = 0.0207 mol CO x- 1 mol CO = 0.0207 mol CH3OH No of moles of CH,OH produced from H, = 0.0317 mol H, X • 2 mol H, = 0.01585 mol CH, OH Since CO imits the production of CH,OH, H, acts as the limiting reactant No of moles of CH,OH formed = 0.01585 mol Amount of CH,OH formed = 0.01585 mol CH,OH 32.0 g 1 mol CH OH = 0.508 g