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How does a balanced equation let us predict the amount of product from a given amount of reactant?

Use mole ratios from a balanced equation to calculate the amounts of reactants and products in mole-to-mole and mass-to-mass problems (MA STE HS-PS1-7(MA), proportional reasoning in reactions).

A standard-level answer on stoichiometric calculations for Massachusetts high school chemistry: reading mole ratios from a balanced equation and using them for mole-to-mole and mass-to-mass calculations through the mole-ratio bridge, grounded in HS-PS1-7(MA).

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  1. What this topic is asking
  2. The mole ratio
  3. The mole-ratio bridge
  4. Mole-to-mole calculations
  5. Mass-to-mass calculations
  6. Try this

What this topic is asking

Standard HS-PS1-7(MA) asks you to use proportional reasoning to solve problems about the quantities of reactants and products. This is stoichiometry: using the coefficients of a balanced equation as a recipe to calculate how much of one substance reacts with or produces another. It is the central calculation skill of the whole course, and everything in this module builds toward it.

The mole ratio

In 2H2+O2→2H2O2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}, the coefficients say that 2 moles of hydrogen react with 1 mole of oxygen to give 2 moles of water. So the mole ratio of hydrogen to water is 2 to 2 (or 1 to 1), and the ratio of oxygen to water is 1 to 2. These ratios let you convert moles of any substance into moles of any other. The equation must be balanced first, because unbalanced coefficients give a false ratio.

The mole-ratio bridge

Every stoichiometry problem follows the same shape, with moles as the bridge between the known and the unknown:

  1. Convert the known to moles. If you are given a mass, divide by molar mass; if a volume of gas at STP, divide by 22.4 L/mol; if a solution, multiply molarity by volume.
  2. Cross the bridge with the mole ratio. Multiply by the ratio of coefficients (unknown over known).
  3. Convert moles of the unknown to the units asked for. Multiply by molar mass for grams, by 22.4 L/mol for liters of gas, and so on.

Mole-to-mole calculations

The simplest case starts and ends in moles, so only the middle step is needed.

Mass-to-mass calculations

The most common exam problem gives a mass and asks for a mass. The path is mass to moles, moles to moles, moles to mass.

Try this

Q1. For 2KClO3→2KCl+3O22\text{KClO}_3 \rightarrow 2\text{KCl} + 3\text{O}_2, how many moles of oxygen form from 4 mol of KClO3\text{KClO}_3? [1]

  • Cue. Ratio of KClO3\text{KClO}_3 to O2\text{O}_2 is 2 to 3, so 4×32=64 \times \dfrac{3}{2} = 6 mol.

Q2. Why must the equation be balanced before you read the mole ratio? [1]

  • Cue. The mole ratio is the ratio of coefficients; unbalanced coefficients do not represent the true reacting proportions, so the calculation would be wrong.

Exam-style practice questions

Practice questions written in the style of MA DESE exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

MA Chemistry (style)3 marksFor N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3: (a) State the mole ratio of hydrogen to ammonia. (b) How many moles of ammonia form from 6 mol of hydrogen? (c) How many moles of nitrogen are needed for 6 mol of hydrogen?
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A 3-point mole-ratio item.

(a) 1 point: the ratio of H2\text{H}_2 to NH3\text{NH}_3 is 3 to 2.
(b) 1 point: 6 mol H2×2 mol NH33 mol H2=46 \text{ mol H}_2 \times \dfrac{2 \text{ mol NH}_3}{3 \text{ mol H}_2} = 4 mol ammonia.
(c) 1 point: 6 mol H2×1 mol N23 mol H2=26 \text{ mol H}_2 \times \dfrac{1 \text{ mol N}_2}{3 \text{ mol H}_2} = 2 mol nitrogen. Markers reward using the coefficients as the conversion ratio.

MA Chemistry (style)3 marksMagnesium burns: 2Mg+O2→2MgO2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}. What mass of magnesium oxide forms from 48 g of magnesium? Use Mg = 24, O = 16 g/mol.
Show worked answer →

A 3-point mass-to-mass item.

1 point: moles of Mg =4824=2.0= \dfrac{48}{24} = 2.0 mol.
1 point: mole ratio of Mg to MgO is 2 to 2, so moles of MgO =2.0= 2.0 mol.
1 point: molar mass of MgO =24+16=40= 24 + 16 = 40 g/mol, so mass =2.0×40=80= 2.0 \times 40 = 80 g. Markers reward the full mass-to-moles-to-moles-to-mass path.

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