Higher Tier: use the equation ratio between two mole amounts
Введено| English | Значение |
|---|---|
| mole ratio/məʊl ˈreɪʃɪəʊ/ | молярное соотношение |
| stoichiometry/ˌstəʊɪkɪˈɒmətri/ | стехиометрия |
What would explain this observation?
- Higher Tier: A balanced equation links amounts of different substances. It cannot be used by multiplying the starting mass directly by the product coefficient.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Use three steps: convert the known mass to moles, apply the balanced coefficient ratio, then convert the target amount to mass. For Mg + 2HCl → MgCl₂ + H₂, one mole Mg requires two moles HCl and forms one mole each of MgCl₂ and H₂. The ratio concerns amounts, while each substance has its own molar mass.
- mole ratio 摩尔比: The ratio of reacting amounts given by balanced equation coefficients; stoichiometry 化学计量: Quantitative relationships between substances in a balanced reaction.
What is the first step when calculating MgO mass from Mg mass?
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Mole ratios - not mass ratios - come from the equation, so mass must first become moles through the molar mass. Multiplying grams by a coefficient mixes units, and formulae never change to balance masses.
Match each technical term to its precise meaning.
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Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- For 2Mg + O₂ → 2MgO, supplied molar masses Mg=24, O₂=32 and MgO=40 $\dfrac{\text{g}}{\text{mol}}$ give mass proportions 48:32:80. These differ from coefficients 2:1:2. A calculation from one reactant assumes sufficient other reactant and complete reaction unless the question supplies a limiting amount. If the known quantity is product mass, work backwards through the same ratio.
- Balance the equation first and underline the known and requested substances. Write a mole row underneath the coefficients, labelling every substance. Avoid rounding a small mole amount early. State the sufficient-reactant and complete-conversion assumptions. Numerical theoretical masses do not include recovery loss, side reactions or reversible-equilibrium effects.
Which two habits make the investigation or model in this case more defensible?
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Stating the conditions and assumptions lets others repeat the work exactly, and recording the supporting evidence lets them check the conclusion. Hiding disagreeing results removes the evidence that could challenge a wrong conclusion, and one measurement can never establish a wider claim.
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: 6.0 g Mg gives 6.0/24=0.25 mol Mg. Its coefficient equals MgO’s coefficient, so theoretical MgO amount=0.25 mol and mass=0.25×40=10.0 g. Oxygen required is 0.125 mol, mass 0.125×32=4.0 g. The calculated total 10.0 g equals 6.0+4.0 g.
Using 2Mg + O₂ → 2MgO, Mg=24 and MgO=40 $\dfrac{\text{g}}{\text{mol}}$, find theoretical MgO mass from 12 g Mg with excess oxygen. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
Using 2Mg + O₂ → 2MgO, Mg=24 and MgO=40 g per mol, find theoretical MgO mass from 12 g Mg with excess oxygen.
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The result is 20 g. Known: 12 g Mg; Mg = 24 g per mol; ratio Mg:MgO = 2:2 = 1:1; MgO = 40 g per mol. Amount Mg = 12/24 = 0.5 mol, so MgO = 0.5 mol. Mass MgO = 0.5 x 40 = 20 g. The product is heavier because it includes oxygen from the air.
Check the conclusion and its limits
- Coefficients do not replace formula masses. A two-to-two coefficient ratio simplifies to one-to-one. A larger product mass can include atoms from another reactant rather than contradict conservation. The mass calculation is a theoretical prediction under stated assumptions, not proof that an experiment will recover every gram.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Equation coefficients are automatically ratios of masses in grams. This claim is false: Coefficients give ratios of amounts in moles. Equal moles of Mg and MgO have different masses (24 versus 40 g), because each particle of MgO also carries an oxygen atom.
Higher Tier: use the equation ratio between two mole amounts: For 2Mg + O₂ → 2MgO, supplied molar masses Mg=24, O₂=32 and MgO=40 $\dfrac{\text{g}}{\text{mol}}$ give mass proportions 48:32:80. These differ from coefficients 2:1:2. A calculation from one reactant assumes sufficient other reactant and complete reaction unless the question supplies a limiting amount. If the known quantity is product mass, work backwards through the same ratio.
Equation coefficients are automatically ratios of masses in grams.
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Coefficients give ratios of amounts in moles. Equal moles of Mg and MgO have different masses (24 versus 40 g), because each particle of MgO also carries an oxygen atom.
The ratio of reacting amounts given by balanced equation coefficients: write the technical term.
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mole ratio means The ratio of reacting amounts given by balanced equation coefficients.