Example 1
How much AgCl forms from 0.0100 mol Ag+ with excess Cl−? Use Ag++Cl−→ AgCl and M(AgCl)=143.5gmol−1.
The ratio is 1:1, so n(AgCl)=0.0100 mol and m=0.0100×143.5=1.435g, or 1.44g to three significant figures.
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Year 11 · SACE Stage 1
Reaction quantities and limiting reagents
Calculate reacting amounts and theoretical precipitate masses.
Method and worked examples
For a stoichiometric calculation, first write a balanced equation. Convert the known mass or solution quantity to moles using n=Mm or n= cV. Use the coefficient ratio to calculate the required amount, then convert to the requested mass, concentration or volume. Coefficients compare moles, so applying them directly to masses is generally incorrect.
When two reactant amounts are given, compare n divided by the corresponding coefficient. The smaller value identifies the limiting reagent and sets the maximum product amount. Assume complete reaction and the stated purity or solubility conditions unless told otherwise. Theoretical yield is a calculated maximum; percentage yield is actual divided by theoretical yield times 100. In precipitation calculations, use the net ionic ratio and the molar mass of the specified dry solid.
Example 1
Example 2
For 2H2+O2→2H2O, use 0.300 mol H2 and 0.100 mol O2. Find the limiting reagent.
Compare 20.300=0.150 with 10.100=0.100. Oxygen is limiting and forms 0.200 mol water.
Watch out: Stoichiometric proportions depend on molar masses and balanced mole ratios.