Question

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 477 mol of CO and 699 mol of H2 are present, which Is the limiting reactant?

How much of the excess reactant remains unchanged
Answer in units of mol.

How much CH3OH is formed?
Answer in unit of mol

Answer 1

The reaction for formation of methanol from carbon monoxide and hydrogen is

`CO(g)+H_(2)(g)-->CH_(3)OH(g)`

As per balanced equation

one mole of CO will react with two moles of Hydrogen to give one mole of methanol

we are provided with

477 mol of CO and 699 mol of H₂

Thus for for 477 moles of CO we need = 2 X 477 mol of H₂ = 954 mol of H₂

However we are provided with only 699 moles of H₂

Thus limiting reagent is H₂

The given moles of H₂ will react with = 0.5 X 699 moles of CO = 349.5 mol CO

The excess of CO [excess reactant] = 477-349.5 = 127.5 mol [unchanged]

From each mole of CO we will get equal moles of methanol

Moles of methanol formed = moles of CO reacted = 127.5 mol