Methanol (CH3OH) is the simplest of the alcohols. It is synthesized by the reaction of hydrogen and carbon monoxide

CO (g) + 2H2 (g) = CH3OH

If 500 mol of CO and 750 mol of H2 are present, which Is the limiting reactant?

By reacting carbon monoxide and hydrogen the formation of methanol takes place, the reaction is,

CO (g) + 2H₂ (g) ⇔CH₃OH (g)

Based on the given reaction, one mole of methanol is obtained by reacting one mole of carbon monoxide (CO) with the two moles of hydrogen (H₂). It is mentioned in the question that for the reaction 500 mol of carbon monoxide and 750 moles of hydrogen are present.

Therefore for 500 moles of carbon monoxide, there is a requirement of 2 * 500 moles of hydrogen, which is equivalent to 1000 moles of hydrogen (H₂). However, only 750 moles of hydrogen is present. Therefore, the limiting reactant in the given case is H₂. The present moles of H₂ will react with 0.5 * 750 moles of CO = 375 mole of CO

The additional or excess concentration of CO, which is the excess reactant will be, 500-375 = 125 moles.