Question

b.
Methanol, CH3OH, which can be used as a fuel, can be formed by the reaction between carbon monoxide and hydrogen as depicted in the following balanced equation. 2H2 + CO ¨ CH3OH Suppose 356 g of CO and 65 g of H2 are mixed and allowed to react.
(i)
Which species is the limiting factor in the overall reaction? Explain the reason for your answer.
(ii)
Calculate the amount of methanol that could, theoretically, be produced.
(iii)
If 352 g of methanol was actually produced, what is the percentage yield for your reaction?

Answer 1

(i) Convert the masses of reactants to moles. 356 g of CO is 356/28 = 12.7 moles. 65 g of H2 is 65/2.016 = 32.2 moles. There are MORE than twice as many moles of H2, compared to CO, so CO is the limiting reactant.
(ii) The number of moles of CH3OH that can be produced is 12.7 . That equals the number of moles of CO available. That mass of CH3OH formed is
12.7*(32.03) = 406.8 g
(iii) 352/406.8 = ___ yield
(convert to %)

Answer 2

(i) Convert the masses of reactants to moles. 356 g of CO is 356/28 = 12.7 moles. 65 g of H2 is 65/2.016 = 32.2 moles. There are MORE than twice as many moles of H2, compared to CO, so CO is the limiting reactant.
(ii) The number of moles of CH3OH that can be produced is 12.7 . That equals the number of moles of CO available. That mass of CH3OH formed is
12.7*(32.03) = 406.8 g
(iii) 352/406.8 = ___ yield
(convert to %)