Question

Carbon monoxide gas reacts with hydrogen gas to form methanol. CO(g)+2H2(g)→CH3OH(g) A 1.45 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 353 mmHg .

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

Express your answer with the appropriate units.

Answer 1

pressure of CO = 232 / 232 + 353

= 0.4 mmHg

partial pressure of H2 = 353 / 232+ 353

= 0.6 mmHg

now calculate the no of moles using PV = nRT

no of moles of CO

0.4 x 1.45 = n x 0.0821 x 305

0.58 = n x 25.04

n = 0.0232 moles

similarly no of moles of H2

0.6 x 1.45 = n x 0.0821 x 305

n = 0.87 / 25.04

= 0.035 moles

according to given balanced reaction one mole of CO required 2 moles of H2

that means 0.0232 moles CO required 2 x 0.0232 moles H2 = 0.046

but we have 0.035 moles that means H2 is limiting agent

theritical yield

from the equation from 2 moles of H2 one mole of CH3OH will form so

from 0.035 moles 0.035/2 moles of MeOH will form = 0.0175 moles

grams of MeOH = no of moles x molar mass

= 0.0175 x 32.04

= 0.56 grams