Apply stoichiometric relationships to reactions involving gases, calculating volumes, masses or amounts of reactants and products using the mole ratio and molar volume or ideal gas equation

What this dot point is asking

QCAA wants you to extend the mole ratio method from Unit 1 stoichiometry to reactions where one or more species is a gas. You should be able to convert any combination of mass, moles and gas volume across a balanced equation, and use molar volume at SLC or the ideal gas equation depending on the conditions specified.

The answer

Gas stoichiometry uses the same four-step mole map as solid-and-aqueous stoichiometry, with one extra conversion at each gas-phase end: between volume of gas and amount in moles.

The four-step mole map

For any stoichiometric problem:

1. Convert the known quantity to moles.
2. Apply the mole ratio from the balanced equation.
3. Convert the answer in moles to the required unit (mass, volume, concentration).
4. Sanity-check units and significant figures.

For gases, "convert to moles" uses either:

• V_m (when at SLC or another standard condition): n = V / V_m.
• PV = nRT (when conditions are specified): n = PV / (RT).

And "convert from moles to volume" reverses:

• V = n x V_m (at standard conditions).
• V = nRT / P (at any other conditions).

Gas-volume ratios at constant T and P (Avogadro's law shortcut)

When all species are gases and all at the same temperature and pressure, the volume ratio of any two species equals the mole ratio. You do not need to convert through moles at all.

Example: in 2H_2(g) + O_2(g) -> 2H_2O(g), 10 L of H_2 reacts with 5 L of O_2 to give 10 L of water vapour, all at the same T and P. This shortcut only works for species in the gas phase under the stated conditions. If water is condensed (liquid), it is not in the volume ratio.

Worked example: mass-to-volume

Calculate the volume of CO_2 produced at SLC when 12.0 g of calcium carbonate decomposes completely.

Equation: CaCO_3(s) -> CaO(s) + CO_2(g).

1. n(CaCO_3) = 12.0 / 100.09 = 0.1199 mol.
2. Mole ratio CaCO_3 : CO_2 = 1 : 1, so n(CO_2) = 0.1199 mol.
3. V(CO_2) = n x V_m = 0.1199 x 24.79 = 2.97 L at SLC.

Worked example: volume-to-mass

What mass of magnesium oxide forms when magnesium is burned in 5.00 L of oxygen gas at 25 degrees C and 100 kPa?

Equation: 2Mg(s) + O_2(g) -> 2MgO(s).

1. n(O_2) = V / V_m = 5.00 / 24.79 = 0.2017 mol (at SLC).
2. Mole ratio O_2 : MgO = 1 : 2, so n(MgO) = 2 x 0.2017 = 0.4034 mol.
3. m(MgO) = n x M = 0.4034 x 40.30 = 16.3 g.

Limiting reactant problems with gases

When two reactants are given (one or both gases), determine the limiting reactant first.

Example: 4.00 L of H_2 reacts with 1.50 L of O_2 at SLC: 2H_2(g) + O_2(g) -> 2H_2O(l). Which is limiting, and what volume of water (liquid, density 1.00 g/mL) forms?

n(H_2) = 4.00 / 24.79 = 0.1614 mol.

n(O_2) = 1.50 / 24.79 = 0.0605 mol.

Stoichiometric ratio needed: H_2 : O_2 = 2 : 1.

Available ratio: 0.1614 / 0.0605 = 2.67. Excess H_2; O_2 is the limiting reactant.

n(H_2O) = 2 x n(O_2) = 2 x 0.0605 = 0.1210 mol.

m(H_2O) = 0.1210 x 18.02 = 2.18 g. Volume = 2.18 mL.

Percent yield with gases

Percent yield = (actual yield / theoretical yield) x 100 percent. Express both yields in the same units (both as volume, both as mass, or both as moles).

Example: a reaction theoretically gives 0.500 mol of NH_3 from a fixed amount of N_2 and H_2 at 200 degrees C and 300 kPa; the actual yield collected is 8.50 L. Find the percent yield.

Theoretical V(NH_3) = nRT / P = (0.500 x 8.314 x 473) / 300 = 1965 / 300 = 6.55 L.

Wait, that gives an actual yield larger than theoretical, which is impossible. Re-examine the data: if the equation predicts 0.500 mol and you measured 8.50 L at the same conditions, your actual moles = PV / (RT) = (300 x 8.50) / (8.314 x 473) = 2550 / 3932.5 = 0.648 mol, which exceeds theoretical and signals either a data error or measurement at the wrong conditions. In a real problem this should ring an alarm. (Always sanity-check that actual is less than or equal to theoretical.)

Reworked with a plausible figure (actual 4.50 L at the same conditions):

n(actual) = (300 x 4.50) / (8.314 x 473) = 1350 / 3932.5 = 0.343 mol.

Percent yield = (0.343 / 0.500) x 100 = 68.6 percent.

When to use what

Examples in context

Example 1. LNG dispatch at Curtis Island, Gladstone. The QCLNG plant on Curtis Island processes coal-seam methane for export. When $1.00 \, \text{tonne}$ of $\text{CH}_4$ combusts during a flare event, the stoichiometry is $\text{CH}_4 + 2 \text{O}_2 \rightarrow \text{CO}_2 + 2 \text{H}_2 \text{O}$. Moles of methane $= 10^6 / 16.05 = 6.23 \times 10^4 \, \text{mol}$, so moles of $\text{CO}_2$ also $6.23 \times 10^4 \, \text{mol}$. At SLC ($25^{\circ}\text{C}$, $100 \, \text{kPa}$) the molar volume is $24.79 \, \text{L mol}^{-1}$, giving a flare $\text{CO}_2$ release of $1.54 \times 10^6 \, \text{L}$ or $1{,}540 \, \text{m}^3$ per tonne methane.

Example 2. Sodium hydroxide carbonation in Brisbane water-treatment. SEQ Water uses $\text{NaOH}$ to raise pH at the Mt Crosby plant. Old caustic stock can react with atmospheric $\text{CO}_2$: $2 \text{NaOH} + \text{CO}_2 \rightarrow \text{Na}_2 \text{CO}_3 + \text{H}_2 \text{O}$. A $5.0 \, \text{L}$ bulk tank loses $4.0 \, \text{g}$ of $\text{NaOH}$ over a month. Moles of $\text{NaOH}$ consumed $= 4.0 / 40.00 = 0.100 \, \text{mol}$; from the $2\!:\!1$ stoichiometry, moles of $\text{CO}_2$ absorbed $= 0.0500 \, \text{mol}$; volume at SLC $= 0.0500 \times 24.79 = 1.24 \, \text{L}$. The operator can use this to size head-space ventilation for IA3 plant-design tasks.

Try this

Q1. State the molar volume of an ideal gas at SLC and at STP. [2 marks]

• Cue. SLC ($25^{\circ}\text{C}$, $100 \, \text{kPa}$): $24.79 \, \text{L mol}^{-1}$. STP ($0^{\circ}\text{C}$, $100 \, \text{kPa}$): $22.71 \, \text{L mol}^{-1}$.

Q2. Calculate the volume of $\text{H}_2$ produced at SLC when $0.486 \, \text{g}$ Mg reacts with excess HCl. [4 marks]

• Cue. $n(\text{Mg}) = 0.0200 \, \text{mol}$; from $\text{Mg} + 2 \text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2$, $n(\text{H}_2) = 0.0200 \, \text{mol}$; $V = 0.496 \, \text{L} = 496 \, \text{mL}$.

Q3. Ammonia is synthesised by $\text{N}_2 + 3 \text{H}_2 \rightarrow 2 \text{NH}_3$. Starting with $50.0 \, \text{L}$ $\text{N}_2$ and $120.0 \, \text{L}$ $\text{H}_2$ at SLC: (a) Identify limiting reagent. (b) Calculate $V(\text{NH}_3)$. (c) State excess volume. [3+2+2 marks]

• Cue. (a) $\text{H}_2$ (need $150 \, \text{L}$). (b) $V(\text{NH}_3) = 80.0 \, \text{L}$. (c) Excess $\text{N}_2 = 10.0 \, \text{L}$. ISMG analysis + judgement.