Consider the following balanced equation : CO (g) + 2H2 (g) ⟶ CH3OH (l) How many moles of CH3OH (l) can be obtained by reacting 2·0 mole of CO (g) with 2·0 mole of H2 (g) ?

Correct — A, (a) 1. This is a limiting reagent calculation, and the balanced equation printed in the stem gives the ratio that decides it: one mole of CO reacts with TWO moles of H2 to give one mole of CH3OH. Test each reactant in turn against that ratio. If all 2·0 mole of CO were to react, it would need 2 x 2·0 = 4·0 mole of hydrogen, but only 2·0 mole of hydrogen has been supplied — so carbon monoxide cannot all react and hydrogen runs out first. Hydrogen is therefore the LIMITING reagent, and it alone fixes the yield: 2·0 mole of H2, divided by the coefficient 2, gives 1·0 mole of CH3OH. Carbon monoxide is the reagent in excess; 1·0 mole of it is consumed and 1·0 mole is left over unreacted at the end. The single most important habit here is never to answer from the reactant that happens to be listed first or to be present in the same amount as the other — the coefficients in the balanced equation, not the raw quantities, decide which reactant limits the reaction.

Why the others are wrong

• (b)2 — Two moles is what you get by ignoring the coefficient 2 on hydrogen and reading the equation as though one mole of CO reacted with one mole of H2 to give one mole of CH3OH. On that misreading the two reactants are supplied in exactly matching amounts, neither limits the other, and 2·0 mole of each yields 2·0 mole of product. It is the commonest wrong answer on any limiting reagent question, because equal starting amounts look like a balanced situation and the eye slides past a coefficient that is printed as part of the formula, 2H2. The equation is printed in the stem precisely so that the coefficient can be checked, and checking it is the whole question.
• (c)3 — There is no route through this equation that produces three moles of methanol from 2·0 mole of carbon monoxide and 2·0 mole of hydrogen — not by taking either reactant as limiting, not by adding the quantities, and not by any misreading of the coefficients. The option is here to complete the sequence 1, 2, 3, 4, so that a candidate who cannot do the calculation has four evenly spaced numbers to guess between rather than a set that gives the answer away by its shape. Its presence is a reminder that a plausible-looking option set is not evidence that every option corresponds to a possible error.
• (d)4 — Four is reachable by two different wrong routes, which is what makes it the best-constructed distractor in the set. The first is simply adding the quantities given — 2·0 mole plus 2·0 mole — and treating the total amount of reactant as the amount of product, which ignores both the stoichiometry and the fact that three moles of gas are consumed for every one mole of liquid produced. The second is subtler: 4·0 is a real number in the correct working, being the quantity of hydrogen that WOULD be needed to consume all 2·0 mole of the carbon monoxide. A candidate who reaches that figure and then writes it down as the answer has done the right calculation and reported the wrong quantity.

Stoichiometry is the arithmetic of a balanced chemical equation, and its central idea is that the coefficients state a fixed MOLE RATIO in which substances react and are formed. In CO + 2H2 giving CH3OH, the ratio is 1 : 2 : 1, so hydrogen is always consumed at twice the rate of carbon monoxide. When two reactants are supplied together they are rarely in exactly that ratio, and the one that runs out first — the LIMITING REAGENT — determines how much product can form; the other is the excess reagent, and whatever is left of it takes no further part. The standard method is mechanical and should be done the same way every time: balance the equation, convert each quantity to moles, divide each reactant's moles by its coefficient, and the smallest quotient identifies the limiting reagent; then use that reactant's mole ratio to the product to get the yield. Here the quotients are 2·0 for carbon monoxide and 1·0 for hydrogen, so hydrogen limits and the yield is 1·0 mole. Note that this is a real industrial reaction, not a textbook invention: methanol is manufactured on a very large scale from synthesis gas, a mixture of carbon monoxide and hydrogen, passed over a catalyst under pressure. Two other conventions in the printed stem are worth reading correctly. The letters in brackets are STATE SYMBOLS — (g) for gas, (l) for liquid, and elsewhere (s) for solid and (aq) for aqueous — and they matter because the reaction converts three moles of gas into one mole of liquid. And the mole itself is a count, not a mass: one mole is Avogadro's number of particles, about 6·022 x 10^23, and the balanced equation counts particles rather than grams.

Key facts

• In CO + 2H2 giving CH3OH, the mole ratio of carbon monoxide to hydrogen to methanol is 1 : 2 : 1.
• The limiting reagent is the reactant that runs out first and determines the maximum yield; the other is the excess reagent.
• To find it, divide each reactant's number of moles by its coefficient in the balanced equation; the smallest quotient identifies the limiting reagent.
• With 2·0 mole of CO and 2·0 mole of H2 the quotients are 2·0 and 1·0, so hydrogen limits the reaction.
• The yield is 1·0 mole of methanol, and 1·0 mole of carbon monoxide is left unreacted.
• Reacting all 2·0 mole of carbon monoxide would have required 4·0 mole of hydrogen.
• One mole is Avogadro's number of particles, about 6·022 x 10^23; a balanced equation counts particles, not grams.
• State symbols in an equation are (s) solid, (l) liquid, (g) gas and (aq) aqueous solution.
• Methanol is manufactured industrially from synthesis gas, a mixture of carbon monoxide and hydrogen, over a catalyst under pressure.

Common traps

• Assuming that equal starting amounts of two reactants means neither is limiting. It depends entirely on the coefficients.
• Reading the 2 in 2H2 as part of the formula rather than as a coefficient, and so using a 1 : 1 ratio.
• Taking the reactant that is named first, or the one with the larger molar mass, as the limiting one without testing.
• Adding the quantities of the two reactants to get the amount of product. Mass is conserved; moles are not.
• Reporting an intermediate figure such as the hydrogen required rather than the product formed.
• Misreading the raised middle dot in 2·0 as a multiplication sign rather than a decimal point.

• practice — not a real PYQ

  For the balanced reaction N2 + 3H2 giving 2NH3, how many moles of ammonia can be obtained by reacting 3 moles of nitrogen with 3 moles of hydrogen ?

  • (a)1
  • (b)2
  • (c)3
  • (d)6

  Answer(b) 2

• practice — not a real PYQ

  In the reaction CO + 2H2 giving CH3OH, 2 moles of CO are reacted with 2 moles of H2. How many moles of CO remain unreacted when the reaction is complete ?

  • (a)0
  • (b)1
  • (c)2
  • (d)4

  Answer(b) 1