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3
54/18 = 3.
Class 12 · English Version · Chemistry
Quantities and chemical calculations
Use moles, concentrations and limiting reagents with explicit units.
Shared cross-level core and stretch unit. Start with the core questions, then use the stretch example for deeper work. The same unit is available in related class and board pathways; it is not a full grade-specific or examination-specification course. A teacher should select the appropriate depth and further topics.
Learn the idea
Amount in moles n = mass/molar mass. Concentration c = n/V uses litres or dm³ when c is in mol/dm³. Equation coefficients give mole ratios, not directly gram ratios. The limiting reagent runs out first according to the balanced equation.
Worked example
For 5.85 g NaCl with molar mass 58.5 g/mol, n = 0.100 mol. In 0.500 dm³ solution, concentration = 0.200 mol/dm³.
- Balance the equation.
- Convert given quantities to moles.
- Apply the ratio, then convert to the requested unit.
Stretch example
For N2 + 3H2 → 2NH3, 2 mol N2 and 3 mol H2 can form at most 2 mol NH3 because H2 limits the reaction. One mole of N2 remains if reaction goes to that extent.
Your written practice
Show your working on paper and enter your answer here. If it does not match, compare the model and try again.
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0.25
Divide by 1000.
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0.4
0.2/0.5 = 0.4.
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2
The coefficient ratio is 1:2.
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Different molar masses mean equal masses contain different amounts in moles.
Use the mole ratio after converting masses.
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1
Only 1 mol N2 reacts with the 3 mol H2.
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Yield = 15/20 × 100 = 75%; some product may have been lost in separation.
State the ratio and a plausible cause; impurities can also distort apparent yield.
Stoichiometry solution
Create a two-reactant calculation and identify its limiting reagent.
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- Use a balanced equation.
- Compare required mole ratios.
- Distinguish theoretical from actual yield.
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