D2.5 Calculate reactant and product quantities from balanced equations

Ontario Grade 11 Chemistry

Quantities in Chemical Reactions

SCH3U D2.5: Use balanced equations to calculate reactant and product quantities

When hydrogen reacts with oxygen, water forms. A balanced equation connects the amounts of substances that react and form. At the particle level, atoms are rearranged into new combinations. At the calculation level, coefficients show the relative amounts involved. This lesson uses those coefficients with mass and mole conversions to find reactant or product quantities.

What you will learn

Explain what coefficients in a balanced equation tell us about amounts of substances.
Use coefficients to form a mole ratio.
Convert between mass and amount in moles to calculate reactant or product quantities.
Keep units through each step and report answers with appropriate significant digits.

1. Begin with the reaction and its particles

A reactant is a starting substance in a chemical reaction. A product is a substance formed by the reaction. You may observe a change when a reaction occurs, but an observation alone does not tell you the quantities of reactants or products.

At the particle level, atoms are rearranged to make new substances. The number of atoms of each element stays the same. A balanced chemical equation represents this by showing equal numbers of each kind of atom on both sides.

A coefficient is the number written in front of a chemical formula. It tells the relative amount of that substance in the reaction. The equation below shows that two amounts of hydrogen react with one amount of oxygen to form two amounts of water. These amounts can be counted as particles or expressed in moles.

Change coefficients to balance an equation. Do not change subscripts. A subscript is the small number within a formula, and it is part of the substance’s identity. Changing it would describe a different substance.

2\mathrm{H_2(g)}+\mathrm{O_2(g)}\rightarrow2\mathrm{H_2O(l)}

Reactants are on the left of the arrow; products are on the right.
Balance the equation before using its coefficients.
Coefficients give relative amounts, not masses.

2. Use coefficients to make a mole ratio

A mole is a counting unit for particles. The symbol

n

stands for amount, and amount is measured in moles, with the unit

\mathrm{mol}

. A mole ratio is a ratio between amounts of two substances. Make it using their coefficients in a balanced equation.

In the hydrogen and oxygen equation, the coefficients show that

2\,\mathrm{mol}

of hydrogen reacts with

1\,\mathrm{mol}

of oxygen. They also show that

2\,\mathrm{mol}

of water forms for every

2\,\mathrm{mol}

of hydrogen that reacts.

To find an unknown amount in moles, start with the known amount and multiply by a ratio made from the equation’s coefficients. Put the substance you want on top and the known substance on the bottom. This lets the known substance’s unit cancel, leaving the unit you need.

Use the balanced equation as written. Its coefficients describe relative amounts, so changing or ignoring them changes the relationship you use.

The general calculation uses the coefficient ratio in an orientation that cancels the known substance and leaves the unknown substance.

n_{\mathrm{unknown}}=n_{\mathrm{known}}\times\frac{c_{\mathrm{unknown}}}{c_{\mathrm{known}}}

Use coefficients from a balanced equation to form the ratio.
Choose the ratio orientation so the known substance cancels.
A mole ratio compares amounts in moles, not masses in grams.

3. Connect mass, moles, and molar mass

A question may give a mass rather than an amount in moles. Molar mass is the mass of one mole of a substance. It is usually written in

\mathrm{g/mol}

. Find it by adding the atomic masses in the substance’s formula, using the periodic table.

To convert a mass to moles, divide the mass by the molar mass. After using the balanced equation to find the amount of another substance, convert that amount to mass if the question asks for grams.

Keep units in each step. In a mass-to-moles conversion, grams cancel and moles remain. In a mole-ratio step, the known substance cancels. In a moles-to-mass conversion, moles cancel and grams remain.

Carry extra digits during the calculation and round the final answer to appropriate significant digits. Significant digits are the meaningful digits in a measured value. The given measurement in a question usually sets the precision of the answer. Keep unrounded values during intermediate steps so rounding does not affect the final result.

n=\frac{m}{M}

Convert mass to moles by dividing by molar mass.
Use the mole ratio to connect amounts of different substances.
Convert moles to mass by multiplying by molar mass.
Show units and round the final answer appropriately.

4. Follow a calculation pathway

First balance the equation. Identify the given quantity and the substance whose quantity is requested. If the given information is a mass, convert it to moles. Then use a coefficient ratio to find the moles of the requested substance. If the question asks for a mass, use that substance’s molar mass to convert the answer to grams.

This sequence connects three ideas: molar mass links mass to moles, coefficients link the amounts of different substances, and molar mass can convert the final amount back to mass. Check that the final unit matches the question.

A balanced equation does not say that the numerical masses of reactants and products are equal for each substance. Instead, it describes relative amounts that react and form. The number of atoms of each element is conserved across the reaction. Your calculation finds the amount of a named substance from the amount of another substance.

m=nM

Balance first, then convert to moles, use the ratio, and convert to the requested unit.
Use the coefficient for each substance as written in the balanced equation.
Check units and significant digits before reporting the result.

Worked example

Find the masses of a reactant and a product

Aluminium reacts with chlorine gas to form aluminium chloride. Calculate the mass of chlorine needed to react with

5.40\,\mathrm{g}

of aluminium and the mass of aluminium chloride formed. Use

M(\mathrm{Al})=26.98\,\mathrm{g/mol}

,

M(\mathrm{Cl_2})=70.90\,\mathrm{g/mol}

, and

M(\mathrm{AlCl_3})=133.33\,\mathrm{g/mol}

. Assume the reaction proceeds as written.

Balance the equation
Balance the equation before using its coefficients. There are two aluminium atoms and six chlorine atoms on each side.
$2\mathrm{Al(s)}+3\mathrm{Cl_2(g)}\rightarrow2\mathrm{AlCl_3(s)}$
Convert aluminium mass to moles
Divide the given aluminium mass by its molar mass. The gram units cancel, leaving moles of aluminium.
$5.40\,\mathrm{g\ Al}\times\frac{1\,\mathrm{mol\ Al}}{26.98\,\mathrm{g\ Al}}=0.200\,\mathrm{mol\ Al}$
Find the chlorine amount
The coefficients give a ratio of three moles of chlorine gas for every two moles of aluminium. The aluminium units cancel, leaving the amount of chlorine gas.
$0.200\,\mathrm{mol\ Al}\times\frac{3\,\mathrm{mol\ Cl_2}}{2\,\mathrm{mol\ Al}}=0.300\,\mathrm{mol\ Cl_2}$
Convert chlorine amount to mass
Multiply the chlorine amount by its molar mass. The result is the mass of chlorine needed, rounded to three significant digits.
$0.300\,\mathrm{mol\ Cl_2}\times70.90\,\mathrm{g/mol}=21.3\,\mathrm{g\ Cl_2}$
Find the aluminium chloride amount
The coefficients give a two-to-two ratio between aluminium and aluminium chloride. Use that ratio to find the product amount in moles.
$0.200\,\mathrm{mol\ Al}\times\frac{2\,\mathrm{mol\ AlCl_3}}{2\,\mathrm{mol\ Al}}=0.200\,\mathrm{mol\ AlCl_3}$
Convert product amount to mass
Multiply the product amount by its molar mass. Round the result to three significant digits, matching the precision of the given aluminium mass.
$0.200\,\mathrm{mol\ AlCl_3}\times133.33\,\mathrm{g/mol}=26.7\,\mathrm{g\ AlCl_3}$

Answer: The reaction requires

21.3\,\mathrm{g}

of chlorine gas and forms

26.7\,\mathrm{g}

of aluminium chloride.

Check: The product mass is approximately the combined reactant masses:

5.40\,\mathrm{g}+21.3\,\mathrm{g}=26.7\,\mathrm{g}

. The balanced equation also has equal atom counts on both sides.

Common mistakes and how to avoid them

Using coefficients from an unbalanced equation.

Correction: Balance the equation first. Only then use the coefficients to form a mole ratio.

Changing a subscript while balancing.

Correction: Change coefficients only. Subscripts are part of a substance’s formula.

Applying a coefficient ratio directly to masses in grams.

Correction: Convert the given mass to moles first. The coefficients compare amounts in moles.

Stopping at moles when the question asks for a mass.

Correction: Convert the final amount in moles to mass using the molar mass of the requested substance.

Dropping units or rounding too early.

Correction: Carry units through each step, keep extra digits during calculations, and round the final answer appropriately.

Lesson summary

A balanced equation conserves atoms and provides coefficients for mole ratios.
Use molar mass to convert between mass and amount in moles.
Use a mole ratio to connect amounts of different substances.
Keep units throughout and report a final answer with appropriate significant digits.

Check your understanding

Question 1

For

2\mathrm{H_2(g)}+\mathrm{O_2(g)}\rightarrow2\mathrm{H_2O(l)}

, how many moles of oxygen gas are needed for

4.0\,\mathrm{mol}

of hydrogen?

$1.0\,\mathrm{mol\ O_2}$
$2.0\,\mathrm{mol\ O_2}$
$4.0\,\mathrm{mol\ O_2}$
$8.0\,\mathrm{mol\ O_2}$

Show answer and explanation

2.0\,\mathrm{mol\ O_2}

The coefficient ratio is one mole of oxygen gas for every two moles of hydrogen. Therefore,

4.0\,\mathrm{mol\ H_2}

corresponds to

2.0\,\mathrm{mol\ O_2}

.

Question 2

For

\mathrm{N_2(g)}+3\mathrm{H_2(g)}\rightarrow2\mathrm{NH_3(g)}

, how many moles of ammonia form from

6.0\,\mathrm{mol}

of hydrogen, if nitrogen is available?

$2.0\,\mathrm{mol\ NH_3}$
$4.0\,\mathrm{mol\ NH_3}$
$6.0\,\mathrm{mol\ NH_3}$
$9.0\,\mathrm{mol\ NH_3}$

Show answer and explanation

4.0\,\mathrm{mol\ NH_3}

The coefficient ratio is two moles of ammonia for every three moles of hydrogen. Thus,

6.0\,\mathrm{mol\ H_2}

corresponds to

4.0\,\mathrm{mol\ NH_3}

.

Key terms

Reactant: A starting substance in a chemical reaction.
Product: A substance formed in a chemical reaction.
Coefficient: A number in front of a formula that shows its relative amount in a balanced equation.
Mole: A counting unit for particles, used as an amount in chemistry calculations.
Mole ratio: A ratio between amounts of substances taken from coefficients in a balanced chemical equation.
Molar mass: The mass of one mole of a substance, usually expressed in grams per mole.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Chemistry (SCH3U), expectation D2.5. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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D2.5 · Calculate reactant and product quantities from balanced equations