Reaction calculations
Introduced| English | Meaning |
|---|---|
| mole ratio/məʊl ˈreɪʃɪəʊ/ | The ratio of amounts of substances measured in moles, often given by the coefficients in a balanced equation. |
| limiting reactant/ˈlɪmɪtɪŋ rɪˈæktənt/ | The reactant that is used up first and therefore limits the amount of product formed. |
| yields/jiːldz/ | The amount of product obtained in a reaction, often expressed as a percentage of the theoretical amount. |
What stops a reaction making more product?
- A balanced equation gives a ratio, but the amount of product also depends on how much of each reactant is available.
- This Supplement lesson predicts the theoretical amount: the maximum for the stated reaction. The amount actually collected can be lower.
Balance the equation
Step the coefficients until every element has the same number of atoms on both sides.
Reaction calculation route
Follow a known mass through a balanced equation to the answer.
Use coefficients as a mole ratio 摩尔比
- In $2\mathrm{H_2}+\mathrm{O_2}\rightarrow2\mathrm{H_2O}$, the mole ratio H₂ : O₂ : H₂O is 2 : 1 : 2. It is not a mass ratio.
- Convert the known quantity to moles, use the ratio to find the required moles, then convert to the requested mass or gas volume.
With excess oxygen, 4.0 g of H₂ has $n=4.0/2=2.0\ \mathrm{mol}$. The ratio gives 2.0 mol of water, so $m=2.0\times18=36\ \mathrm{g}$. Oxygen contributes mass too: the product can weigh more than the hydrogen alone.
How many moles are in 4 g of hydrogen gas (Mr = 2)?
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moles = 4 ÷ 2 = 2.
In 2H₂ + O₂ → 2H₂O, the mole ratio of H₂ to H₂O is:
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2 moles of H₂ give 2 moles of water.
If 2 moles of water form (Mr of water = 18), what mass of water is that?
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mass = 2 × 18 = 36 g.
Put the reaction-calculation steps in order.
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Mass → moles → ratio → moles → mass.
The big numbers in a balanced equation give the ______ ratio.
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Reactions combine in whole-number mole ratios.
N₂ + 3H₂ → 2NH₃. With nitrogen in excess, how many moles of NH₃ form from 0.60 mol H₂?
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The ratio H₂:NH₃ is 3:2. n(NH₃)=0.60×2/3=0.40 mol.
Find the limiting reactant 限量反应物
- The limiting reactant runs out first and sets the theoretical product amount. A reactant in excess remains after the limiting reactant is used up.
- For each reactant, calculate moles ÷ its equation coefficient. The smaller result sets how far the reaction can proceed. Compare moles with the ratio, not raw masses or raw mole counts.
This model uses a 1 : 1 ratio; use the actual equation's ratio for each calculation.
Work with two measured reactants
- Mix 6.0 g H₂ and 32 g O₂. Their amounts are 3.0 mol and 1.0 mol. Compare $3.0/2=1.5$ with $1.0/1=1.0$: oxygen limits the reaction.
- One mole of O₂ uses 2.0 mol H₂ and forms 2.0 mol water. Water mass $=2.0\times18=36\ \mathrm{g}$. Hydrogen left $=(3.0-2.0)\times2=2.0\ \mathrm{g}$; total final mass $36+2=38\ \mathrm{g}$ matches the initial $6+32$.
Do not apply a ratio until the equation is balanced. Do not assume that the reactant with the smaller mass, or even the smaller number of moles, always limits the reaction.
Mix 2.0 mol H₂ and 2.0 mol O₂. For 2H₂ + O₂ → 2H₂O, which is limiting?
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Two moles H₂ need only one mole O₂. Hydrogen runs out first; one mole oxygen remains.
Mix 2.0 g H₂ and 8.0 g O₂. Use 2H₂ + O₂ → 2H₂O and molar masses H₂=2, O₂=32, H₂O=18 $\dfrac{\mathrm{g}}{\mathrm{mol}}$. Find the theoretical mass of water in g.
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Amounts are 1.0 mol H₂ and 0.25 mol O₂. Oxygen limits, making 2×0.25=0.50 mol water; mass=0.50×18=9.0 g.
Convert product moles to gas volume
- For $\mathrm{CaCO_3}\rightarrow\mathrm{CaO}+\mathrm{CO_2}$, one mole of carbonate makes one mole of carbon dioxide. A 10.0 g pure carbonate sample has $n=10.0/100=0.100\ \mathrm{mol}$.
- If it decomposes completely, it makes 0.100 mol CO₂: $V=0.100\times24=2.40\ \mathrm{dm^3}$ at r.t.p. A mass of unreactive impurity would not make CO₂.
CaCO₃ → CaO + CO₂. Find the CO₂ volume at r.t.p. from complete decomposition of 5.0 g pure CaCO₃. Its molar mass is 100 $\dfrac{\mathrm{g}}{\mathrm{mol}}$; use 24 dm³ per mol. Answer in dm³.
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n(CaCO₃)=5.0/100=0.050 mol; n(CO₂)=0.050 mol, so V=0.050×24=1.20 dm³.
Predict and check a new reaction
- Read whether another reactant is in excess or whether both amounts are given. Find the limiting reactant when necessary.
- Give the theoretical result with units and state the assumption of complete reaction. Later lessons compare predicted and collected yields 产率 and investigate impure samples.
Known quantity → moles → balanced ratio → required moles → requested quantity. The limiting reactant controls the result; excess material does not all turn into product.
CaCO₃ → CaO + CO₂. Exactly 2.0 mol pure CaCO₃ decomposes completely. Find CO₂ mass using molar mass 44 $\dfrac{\mathrm{g}}{\mathrm{mol}}$ and show why the equation matters.
A relative mass alone is not the product mass: first find the product amount.
For CaCO₃ → CaO + CO₂, 2.0 mol CaCO₃ decomposes completely. Molar mass CO₂=44 $\dfrac{\mathrm{g}}{\mathrm{mol}}$. What CO₂ mass forms?
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The equation gives one mole of CO2 per mole of carbonate.
For 2H₂ + O₂ → 2H₂O, 1.0 mol H₂ is mixed with 0.25 mol O₂. What amount of H₂ remains after complete reaction, in mol?
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0.25 mol O₂ consumes 0.50 mol H₂. Remaining H₂=1.0−0.50=0.50 mol.
For 2Mg + O₂ → 2MgO, a learner says 2 g magnesium reacts with 1 g oxygen because the coefficients are 2:1. Explain the mistake. Use Ar: Mg=24 and O=16.
Compare the reasoning and each criterion; equivalent wording is valid.