How much can we make?

The mole map

Every calculation goes through moles: n = m ÷ M for solids and liquids, n = cV for solutions (V in litres), n = V ÷ 22.71 for gases at STP (0.0 °C and 100.0 kPa), and n = PV ÷ RT for gases at any other conditions, with P in kPa, V in L, T in kelvin and R = 8.314 J K⁻¹ mol⁻¹.

Limiting and excess reagents

The limiting reagent is the reactant that runs out first; it fixes how much product can form. The other reactant is in excess. You cannot tell which is limiting from the masses: convert to moles and compare with the equation’s ratio. 10.0 g of N₂ (0.357 mol) needs 1.07 mol of H₂, so 3.00 g of H₂ (1.49 mol) is in excess even though its mass is smaller.

Try it with a real experiment: magnesium ribbon dropped into hydrochloric acid, with the hydrogen collected in a gas syringe. Change the amounts and see which reagent limits the volume.

Notice that once magnesium is limiting, adding more acid does not produce more gas; it only makes the reaction faster.

Theoretical and percentage yield

The theoretical yield is the maximum amount of product, calculated from the limiting reagent. The percentage yield = actual ÷ theoretical × 100. Yields fall below 100% because reactions may be reversible, side reactions form other products, and product is lost when it is filtered, washed, recrystallised or transferred. In a synthesis with several steps, the overall yield is the product of the step yields: 85% × 70% × 60% is only 35.7%.

Gases at industrial conditions

Industrial gases are rarely at STP. Roasting 2.20 tonnes of zinc sulfide with a 72.0% yield gives 1.625 × 10⁴ mol of SO₂; at 300.0 °C (573.15 K) and 165.0 kPa this occupies V = nRT ÷ P = 4.69 × 10⁵ L. The answer is given to 3 significant figures because the least precise data (2.20 t, 72.0%) have 3.