Gas Stoichiometry气体计量

Big Idea · Transformations (TRA)AP Science Practice 5 · Mathematical RoutinesAP link · Unit 4~40 min

Learning goals

  • Convert between gas volume and moles in stoichiometry problems, using molar volume at STP or PV = nRT
  • Use the fact that, at the same T and P, gas volumes react in the ratio of their coefficients
  • Recognize which substances can and cannot use 22.4 L/mol

Before you start

  • Stoichiometry (mole ratios) (2.6.5)
  • The ideal gas law PV = nRT (2.8.4)

An airbag inflates in 30 milliseconds

In a serious crash, a car’s airbag has to fill in about 30 milliseconds. It contains sodium azide, NaN₃, which the impact sets off: it decomposes almost instantly, releasing a large volume of nitrogen gas. How many grams of NaN₃ does it take to fill a 60-L airbag? That is a classic gas stoichiometry problem: the given quantity is a gas volume, and the wanted quantity is the mass of a solid.

Gas volumes at both ends of the mole bridge

The stoichiometry of 2.6.5 runs mass → moles → moles → mass. Moles are the bridge in the middle, and either end can be anything that converts to moles. For a gas, there are two ways to convert between volume and moles:

  • At STP (0 °C, 1 atm): n = V ÷ 22.4 L/mol.
  • At any temperature and pressure: n = PV ÷ RT.

So the full route is: given (mass or gas volume) → moles A → mole ratio → moles B → wanted (mass or gas volume).

Gas volumes react in the ratio of the coefficients

By Avogadro’s law, gas volume is proportional to moles at the same temperature and pressure. So for the gases in a reaction:

At the same temperature and pressure, the volumes of reacting and produced gases are in the same ratio as their coefficients.

For 2H₂(g) + O₂(g) → 2H₂O(g), 2 L of H₂ reacts with 1 L of O₂ to make 2 L of water vapor, all at the same T and P. Gay-Lussac discovered this pattern in 1808; it is called Gay-Lussac’s law of combining volumes气体反应体积定律, and it later helped Avogadro argue that gases are made of molecules. It is a different law from the Gay-Lussac’s law (P ∝ T) in 2.8.2.

The rule works only for gases at the same T and P. Liquids and solids don’t follow it; if the water produced is liquid, its volume can’t be found from the coefficients.

This lesson uses the atomic masses on the AP periodic table (H 1.008, C 12.01, N 14.01, O 16.00, Na 22.99, Cl 35.45, K 39.10, Ca 40.08, Zn 65.38).

Pick a reaction, choose mass or gas volume for both the given and the wanted quantity, and use STP or your own temperature and pressure:

Try it: gas stoichiometry (mass ⇄ moles ⇄ gas volume)
Given:
Find:
Conditions:
Given amount

    Example 1: the airbag

    The reaction in an airbag is

    2 NaNX3(s)→2 Na(s)+3 NX2(g)\ce{2NaN3(s) -> 2Na(s) + 3N2(g)}

    What mass of NaN₃ is needed to produce 60.0 L of N₂ at 25 °C and 1.00 atm?

    Step-by-step solution

    1. The route. V(N₂) → mol N₂ → mol NaN₃ → g NaN₃. Moles are the middle stop, because coefficients give mole ratios.

    2. Gas volume → moles. 25 °C is not STP, so 22.4 L/mol doesn’t apply; use PV = nRT:

      n(NX2)=PVRT=1.00×60.00.08206×298=2.454 mol\begin{aligned} n(\ce{N2}) &= \frac{PV}{RT} \\ &= \frac{1.00 \times 60.0}{0.08206 \times 298} \\ &= 2.454\ \text{mol} \end{aligned}

      Keep an extra digit for now.

    3. Mole ratio. NaN₃ : N₂ = 2 : 3:

      n(NaNX3)=2.454×23=1.636 mol\begin{aligned} n(\ce{NaN3}) &= 2.454 \times \frac{2}{3} \\ &= 1.636\ \text{mol} \end{aligned}

    4. Moles → mass. M(NaN₃) = 22.99 + 3(14.01) = 65.02 g/mol:

      m=1.636×65.02=106 g\begin{aligned} m &= 1.636 \times 65.02 \\ &= \boxed{106\ \text{g}} \end{aligned}

    5. Think about it. About 100 g of solid releases 60 L of gas, an expansion of hundreds of times, which is why the bag fills in an instant.

    6. A common mistake. Using 22.4 L/mol anyway: 60.0 ÷ 22.4 = 2.68 mol, which leads to 116 g, about 9% too much.

    Example 2: zinc and hydrochloric acid

    13.0 g of zinc reacts with excess hydrochloric acid. What volume of H₂ forms at STP? If all that H₂ is then burned, what volume of O₂ does it need at the same conditions?

    Step-by-step solution

    1. Mass → moles. The acid is in excess, so all the zinc reacts:

      n(Zn)=13.0 g65.38 g/mol=0.1988 mol\begin{aligned} n(\ce{Zn}) &= \frac{13.0\ \text{g}}{65.38\ \text{g/mol}} \\ &= 0.1988\ \text{mol} \end{aligned}

    2. Mole ratio. Zn and H₂ both have a coefficient of 1:

      Zn(s)+2 HCl(aq)→ZnClX2(aq)+HX2(g)\begin{aligned} &\ce{Zn(s) + 2HCl(aq)} \\ &\ce{-> ZnCl2(aq) + H2(g)} \end{aligned}

      so 0.1988 mol H₂ forms.

    3. Moles → volume. At STP, 1 mol of gas is 22.4 L:

      V(HX2)=0.1988×22.4=4.45 L\begin{aligned} V(\ce{H2}) &= 0.1988 \times 22.4 \\ &= \boxed{4.45\ \text{L}} \end{aligned}

    4. Combining volumes. 2H₂(g) + O₂(g) → 2H₂O(l). At the same T and P, gas volumes are in the same ratio as moles, which is the coefficient ratio:

      V(OX2)=4.454×12=2.23 L\begin{aligned} V(\ce{O2}) &= 4.454 \times \frac{1}{2} \\ &= \boxed{2.23\ \text{L}} \end{aligned}

      No need to convert back to moles.

    5. Why this works. By Avogadro’s law, at the same T and P a gas’s volume is proportional to its moles. Use volume ratios only for gases compared at the same T and P.

    Practice: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g), all at the same T and P. What volume of the requested gas?
    1. O₂ needed to burn 1 L of C₃H₈
    2. CO₂ produced from 1 L of C₃H₈
    3. Water vapor produced from 1 L of C₃H₈
    4. O₂ needed to burn 2 L of C₃H₈
    5. C₃H₈ burned by 15 L of O₂
    6. CO₂ produced when exactly 5 L of O₂ reacts
    7. O₂ left over after 1 L of C₃H₈ burns in 10 L of O₂
    8. Water vapor formed along with 3 L of CO₂

    Common mistakes

    Mistake 1: “18 g of water takes up 22.4 L at STP.” At 0 °C, water is a liquid (or ice); 18 g of it is only about 18 mL. 22.4 L/mol applies only to gases.

    Mistake 2: “Gas volumes react in the ratio of their masses.” At the same T and P, gas volumes are in the ratio of their moles, which is the coefficient ratio, whatever the masses. 2 L of light H₂ reacts with 1 L of much heavier O₂.

    Mistake 3: “22.4 L/mol works at any conditions.” At 25 °C a mole of gas is about 24.5 L. When the conditions aren’t 0 °C and 1 atm, convert with PV = nRT.

    Quick check

    6.5 g of zinc reacts with excess hydrochloric acid. What volume of H₂ forms at STP?

    At the same T and P, 10 mL of H₂ and 10 mL of O₂ are mixed and ignited. After the reaction, the mixture returns to the original conditions (water is liquid). What gas remains?

    Summary

    • Gas stoichiometry: given → moles → mole ratio → moles → wanted. Convert gas volume and moles with 22.4 L/mol at STP or with PV = nRT at other conditions.
    • At the same T and P, reacting and produced gas volumes are in the ratio of the coefficients, so you can work with volumes directly.
    • 22.4 L/mol applies only to gases at STP; liquid and solid volumes don’t follow the coefficient ratio.
    On to AP (for later)

    In AP Chemistry Unit 4, stoichiometry problems often have a gas at one end: given the pressure and volume of a gas produced, find the mass of a reactant, or the reverse. AP problems usually state the temperature and pressure and expect PV = nRT rather than 22.4 L/mol. Sometimes the same problem adds collection over water or a limiting reactant.

    Practice this chapter: 2.8 · Gases · 10 questions

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