Calculations from Chemical Equations根据化学方程式计算

Big Idea · Scale, Proportion, and Quantity (SPQ)AP link · Unit 4~25 min

Learning goals

  • Explain what makes calculations from an equation possible
  • Solve problems in six clear steps
  • Use the mass of one substance to find the mass of another

Before you start

  • Relative formula mass (1.5.5)
  • Reading a chemical equation (1.6.2)
  • Balancing chemical equations (1.6.3)

Why calculate?

Factories that make fertilizer, steel, or medicine all need to plan ahead: how much raw material do they need, and how much product will they get? A chemical equation tells you not only what forms, but also how much.

What makes it work

As you saw last lesson, the mass ratio of the substances in an equation is fixed: it equals the ratio of coefficient × formula mass. Take the electrolysis of water:

2 HX2O(l)→electrolysis2 HX2(g)+OX2(g)\ce{2H2O(l) ->[\text{electrolysis}] 2H2(g) + O2(g)}

The mass ratio of water, hydrogen, and oxygen is 36 : 4 : 32, no matter how much water you split. So if you know the mass of one substance, you can find the mass of another with a proportion.

🔧 Math tool card · Basic AlgebraSkip it if you already know this

Solving a proportion: if ab=cx\frac{a}{b} = \frac{c}{x}, cross-multiply to get a⋅x=b⋅ca \cdot x = b \cdot c, so x=b⋅cax = \frac{b \cdot c}{a}.

Example: 25=6x\frac{2}{5} = \frac{6}{x}. Cross-multiplying gives 2x=302x = 30, so x=15x = 15.

Solve 364=18x\frac{36}{4} = \frac{18}{x}. What is x?

Six steps to a solution

  1. Define: let x be the unknown mass (don’t attach a unit to x itself).
  2. Write: write the correct, balanced equation.
  3. Find: under the substances involved, write coefficient × formula mass, then the known mass and x.
  4. Set up: write the proportion.
  5. Solve: solve for x.
  6. Answer: state the answer clearly.

Example 1: Splitting water

What is the maximum mass of hydrogen you can get by electrolyzing 18 g of water? (H = 1, O = 16)

Step-by-step solution
  1. Define. Let x be the mass of hydrogen produced.
  2. Write. 2 HX2O(l)→electrolysis2 HX2(g)+OX2(g)\ce{2H2O(l) ->[\text{electrolysis}] 2H2(g) + O2(g)}
  3. Find. Under 2 HX2O\ce{2H2O}, write 2 × 18 = 36 and 18 g. Under 2 HX2\ce{2H2}, write 2 × 2 = 4 and x.
  4. Set up. 364=18 gx\frac{36}{4} = \frac{18\,\text{g}}{x}
  5. Solve. x=4×18 g36=2 gx = \frac{4 \times 18\,\text{g}}{36} = 2\,\text{g}
  6. Answer. At most 2 g of hydrogen is produced.

Try it

Pick a different given substance, change the mass, or solve for oxygen instead, and watch all six steps update.

Try it: change the given values and watch the solution update
2 HX2O→electrolysis2 HX2+OX2\ce{2H2O ->[\text{electrolysis}] 2H2 + O2}

    Example 2: Heating limestone

    Industry makes quicklime (CaO\ce{CaO}) by strongly heating limestone, which is mostly CaCOX3\ce{CaCO3}. How many tonnes of calcium carbonate are needed to make 5.6 t of calcium oxide? (Ca = 40, C = 12, O = 16)

    Step-by-step solution
    1. Define. Let x be the mass of calcium carbonate needed.
    2. Write. CaCOX3(s)→ΔCaO(s)+COX2(g)\ce{CaCO3(s) ->[\Delta] CaO(s) + CO2(g)}
    3. Find. Under CaCOX3\ce{CaCO3}, write 40 + 12 + 3 × 16 = 100 and x. Under CaO\ce{CaO}, write 40 + 16 = 56 and 5.6 t.
    4. Set up. 10056=x5.6 t\frac{100}{56} = \frac{x}{5.6\,\text{t}}
    5. Solve. x=100×5.6 t56=10 tx = \frac{100 \times 5.6\,\text{t}}{56} = 10\,\text{t}
    6. Answer. 10 t of calcium carbonate is needed. The unit doesn’t have to be grams, as long as the known mass and the answer use the same unit.

    Common mistakes

    Mistake 1: Calculating with an unbalanced equation. If you use the unbalanced HX2O→HX2+OX2\ce{H2O -> H2 + O2}, the mass ratio will be wrong. Always balance the equation in the “Write” step.

    Mistake 2: Forgetting the coefficient. The mass that goes with 2 HX2O\ce{2H2O} is 2 × 18 = 36, not 18.

    Mistake 3: Plugging in the mass of a mixture. The mass in your proportion must be of a pure substance that actually reacts. For 100 t of limestone that is 80% CaCOX3\ce{CaCO3}, use 80 t.

    Quick check

    What is the maximum mass of oxygen from heating 31.6 g of potassium permanganate? (2 KMnOX4(s)→ΔKX2MnOX4(s)+MnOX2(s)+OX2(g)\ce{2KMnO4(s) ->[\Delta] K2MnO4(s) + MnO2(s) + O2(g)}; the formula mass of KMnOX4\ce{KMnO4} is 158)

    3 g of carbon burns completely in plenty of oxygen. How much carbon dioxide forms? (C = 12, O = 16)

    Summary

    • The basis: the mass ratio of the substances in an equation is fixed, equal to the ratio of coefficient × formula mass.
    • The steps: define, write, find, set up, solve, answer.
    • Balance the equation first, multiply formula masses by their coefficients, and plug in only the mass of a pure substance that actually reacts.
    On to AP (for later)

    In Stage 2 you’ll learn about the mole, and calculations will follow the path mass → moles → mole ratio from the coefficients → mass. It’s the same idea as today’s proportions, and it’s the stoichiometry of AP Unit 4 (topic 4.5).

    Practice this chapter: Module 6 · Chemical Equations · 12 questions

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