F. Gases and Atmospheric Chemistry · Gases and Atmospheric Chemistry

Gas Stoichiometry

F2.4 F2.3About 70 minutes

Learning goals

I can use a balanced equation and the mole ratio to relate amounts of gases and other substances.
I can convert between gas volume and moles with PV = nRT or molar volume.
I can use volume ratios directly for gases at the same temperature and pressure.
I can identify a limiting reactant in a reaction between gases.
I can calculate the CO₂ produced by burning a fuel.

Words to know

stoichiometry
The calculation of quantities of reactants and products using a balanced chemical equation.

mole ratio
The ratio of coefficients in a balanced equation, used to convert moles of one substance to moles of another.

gas stoichiometry
Stoichiometry in which at least one quantity is a gas volume, using PV = nRT or molar volume.

limiting reactant
The reactant that is used up first and so limits the amount of product.

excess reactant
A reactant that is left over when the limiting reactant is used up.

How much gas will a reaction make?

A car's airbag must inflate to just the right size in a fraction of a second. Older airbag designs used a solid, sodium azide (NaN₃), which decomposes very quickly when triggered: 2 NaN₃(s) → 2 Na(s) + 3 N₂(g). Engineers need to know exactly how many grams of solid make the right volume of nitrogen. That's gas stoichiometry: using a balanced equation to connect amounts of gases with amounts of other substances.

Remember from Quantities in Chemical Reactions

In the unit Quantities in Chemical Reactions you used the mole ratio from a balanced equation to convert moles of one substance into moles of another, and n = m ÷ M to convert between mass and moles. You also found limiting reactants. Gas stoichiometry adds just one new step: converting between moles and gas volume, using PV = nRT or molar volume.

The road map

Write the balanced equation.
Convert what you're given into moles:

from mass: n = m ÷ M
from a gas volume: n = PV ÷ RT (any conditions), or n = V ÷ Vₘ (only at STP or SATP)

Use the mole ratio to find moles of the substance you want.
Convert moles to what's asked for: mass (m = n × M) or gas volume (V = nRT ÷ P, or V = n × Vₘ).

Shortcut for gases only: if both substances are gases at the same temperature and pressure, the volume ratio equals the mole ratio (Avogadro). You can go straight from litres to litres.

The gas stoichiometry road map. Every route passes through moles, and the mole ratio from the balanced equation is the bridge from substance A to substance B.

Worked example

Example 1: Mass of solid → volume of gas

What volume of nitrogen gas, at 25 °C and 101 kPa, is produced when 130 g of sodium azide decomposes? 2 NaN₃(s) → 2 Na(s) + 3 N₂(g)

Moles of NaN₃: M = 22.99 + 3(14.01) = 65.02 g/mol; n = 130 g ÷ 65.02 g/mol = 2.00 mol
Mole ratio: n(N₂) = 2.00 mol NaN₃ × (3 mol N₂ ÷ 2 mol NaN₃) = 3.00 mol
Volume: V = nRT ÷ P = (3.00)(8.314)(298) ÷ 101 = 73.6 L

73.6 L of N₂

Worked example

Example 2: Volume of gas → mass of fuel

A propane barbecue produces 50.0 L of CO₂, measured at SATP. What mass of propane burned? C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)

Moles of CO₂ (at SATP): n = 50.0 L ÷ 24.8 L/mol = 2.02 mol
Mole ratio: n(C₃H₈) = 2.02 mol × (1 ÷ 3) = 0.672 mol
M(C₃H₈) = 3(12.01) + 8(1.01) = 44.11 g/mol; m = 0.672 × 44.11 = 29.6 g

29.6 g of propane

Worked example

Example 3: A lab reaction

How many millilitres of hydrogen, at 22 °C and 99.5 kPa, are produced when 0.250 g of magnesium reacts with excess hydrochloric acid? Mg(s) + 2 HCl(aq) → MgCl₂(aq) + H₂(g)

n(Mg) = 0.250 g ÷ 24.31 g/mol = 0.01028 mol
Mole ratio 1 : 1, so n(H₂) = 0.01028 mol
V = nRT ÷ P = (0.01028)(8.314)(295) ÷ 99.5 = 0.254 L

254 mL of H₂

Limiting reactants with gases

If two gases react at the same temperature and pressure, you can find the limiting reactant directly from their volumes. Carbon monoxide burns in oxygen: 2 CO(g) + O₂(g) → 2 CO₂(g). Mix 10 L of CO with 10 L of O₂: the 10 L of CO needs only 5 L of O₂, so CO is limiting. You get 10 L of CO₂ and 5 L of O₂ is left over.

Volumes of reacting gases follow the mole ratio. Here oxygen is in excess, so 5 L of it is left over after all the carbon monoxide has reacted.

Worked example

Example 4: Carbon dioxide from a litre of gasoline

Treat gasoline as pure octane, C₈H₁₈ (density 0.703 g/mL). What mass and what volume (at SATP) of CO₂ form when 1.00 L of it burns completely? 2 C₈H₁₈(l) + 25 O₂(g) → 16 CO₂(g) + 18 H₂O(g)

Mass of octane: 1000 mL × 0.703 g/mL = 703 g
M(C₈H₁₈) = 8(12.01) + 18(1.01) = 114.26 g/mol; n = 703 ÷ 114.26 = 6.15 mol
Mole ratio 16 : 2 = 8 : 1, so n(CO₂) = 6.15 × 8 = 49.2 mol
Mass: 49.2 mol × 44.01 g/mol = 2.17 × 10³ g = 2.17 kg
Volume at SATP: 49.2 mol × 24.8 L/mol = 1.22 × 10³ L

About 2.2 kg of CO₂, which would occupy about 1220 L at SATP: a litre of fuel becomes over a thousand litres of greenhouse gas. (Natural Resources Canada uses about 2.3 kg of CO₂ per litre for real gasoline, which is a mixture of hydrocarbons.) You'll use this in the lesson on air quality and carbon footprints.

Hands-on activity

Design a mini airbag

Goal: calculate how much baking soda will inflate a 1 L zip-top sandwich bag with carbon dioxide, then test your prediction.

You need: a grown-up, a 1 L zip-top bag (check that it holds about 1 L of water without bursting), baking soda (sodium hydrogen carbonate, NaHCO₃), white vinegar (5 % acetic acid), a kitchen scale (or measuring spoons: 1 teaspoon of baking soda is about 4 to 5 g), a tissue, a measuring cup, eye protection, and a sink or tray.

Safety: wear eye protection: an over-filled bag can pop and spray vinegar. Work over a sink. Do not use a larger amount than you calculate.

The reaction: NaHCO₃(s) + CH₃COOH(aq) → NaCH₃COO(aq) + H₂O(l) + CO₂(g)

Calculate: assume 22 °C and 101 kPa. n(CO₂) = PV ÷ RT for V = 1.0 L. By the 1 : 1 ratio, that's the moles of NaHCO₃ needed. Convert to grams (M = 84.01 g/mol).
Vinegar contains about 0.83 mol/L of acetic acid. Calculate the volume of vinegar that contains the same number of moles, then use a bit more so the vinegar is in excess and the baking soda is limiting.
Wrap the baking soda in a tissue packet (to delay the reaction). Pour the vinegar into the bag, then drop in the packet and quickly seal the bag. Shake gently.
Was the bag just full? Over-full? Under-full?

What to look for: n ≈ 0.041 mol, so about 3.5 g of baking soda and about 50 mL of vinegar (use about 60 to 70 mL). The bag usually ends up a bit under-filled, because some CO₂ dissolves in the liquid and the reaction may not finish. Engineers face the same problem: a design calculation is a starting point that must be tested.

Just for fun

Sodium azide airbags made a fluffy pillow of plain nitrogen gas, the same gas that makes up about 78 % of the air you're breathing right now. The chemistry was so fast that the bag filled before you could finish saying "stoichiometry".

Your turn

List the four steps for solving a gas stoichiometry problem.

What volume of hydrogen at SATP is produced when 10.0 g of zinc reacts with excess acid? Zn(s) + 2 HCl(aq) → ZnCl₂(aq) + H₂(g)

What volume of CO₂ at STP is released when 3.00 g of calcium carbonate is heated? CaCO₃(s) → CaO(s) + CO₂(g)

What mass of hydrogen peroxide must decompose to give 1.00 L of oxygen at SATP? 2 H₂O₂(aq) → 2 H₂O(l) + O₂(g)

What volume of oxygen is needed to burn 5.0 L of methane, measured at the same T and P? CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g)

How many millilitres of H₂ at 25 °C and 100.0 kPa form when 0.120 g of Mg reacts with excess HCl?

More practice: warm-up

In N₂ + 3 H₂ → 2 NH₃, what is the mole ratio of N₂ to NH₃?

How many moles are in 44.8 L of a gas at STP?

What volume does 0.50 mol of gas occupy at SATP?

For 2 CO(g) + O₂(g) → 2 CO₂(g), how many litres of O₂ react with 6.0 L of CO (same T and P)?

How many moles are in 8.00 g of O₂?

More practice: on your own

What mass of sodium azide is needed to make 60.0 L of N₂ at 25 °C and 101 kPa? 2 NaN₃ → 2 Na + 3 N₂

In photosynthesis, 6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂. What volume of O₂ at SATP is released when a plant makes 18.0 g of glucose?

Electrolysis splits water: 2 H₂O(l) → 2 H₂(g) + O₂(g). What volumes of H₂ and O₂ at SATP come from 12.0 g of water?

What volume of oxygen at SATP is needed to burn 1.00 mol of octane? 2 C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O

15 L of H₂ and 10 L of Cl₂ react: H₂(g) + Cl₂(g) → 2 HCl(g) (same T and P). Which is limiting? What volume of HCl forms, and what is left over?

Challenge

0.0500 g of Mg reacts with excess HCl. The H₂ is collected over water at 21 °C with atmospheric pressure 100.5 kPa. Predict the volume collected.

What mass of potassium chlorate must decompose to give 500 mL of O₂ at 20 °C and 98.0 kPa? 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)

A family car burns 1500 L of gasoline in a year. Using the 2.3 kg CO₂ per litre figure, how many tonnes of CO₂ is that, and what volume would it occupy at SATP?

Check your understanding

Quiz

In gas stoichiometry, the step that connects substance A to substance B is:

For CH₄(g) + 2 O₂(g) → CO₂(g) + 2 H₂O(g), 3.0 L of CH₄ produces how much CO₂ (same T and P)?

2.00 mol of a gaseous product is made at SATP. Its volume is:

When can you use the volume ratio directly as the mole ratio?

20 L of H₂ and 20 L of O₂ react: 2 H₂ + O₂ → 2 H₂O(g). The limiting reactant is:

To convert a gas volume measured at 30 °C and 95 kPa into moles, you should use:

For parents and teachers

Big idea (F2.4, F2.3): gas stoichiometry uses a balanced equation and the mole ratio, with PV = nRT or molar volume to convert between moles and gas volume. It builds directly on the unit Quantities in Chemical Reactions.

How to help: ask your teen to say each step aloud ("grams to moles, mole ratio, moles to litres") before calculating. The road-map diagram is worth copying onto a study card.

Common difficulties:

Skipping the mole ratio (converting grams of A straight to litres of B).
Using 22.4 L/mol at the wrong conditions. Only at STP.
Unbalanced equations. Check the balancing before anything else.

Safety: the baking-soda "airbag" can pop; use eye protection, work over a sink and don't exceed the calculated amounts.