Solve solution-stoichiometry problems

Ontario Grade 11 Chemistry

Solutions and Solubility

Using solution concentration and balanced equations to connect amounts of reactants and products

When two clear solutions are mixed, a solid may appear even though neither solution looked solid before mixing. The solid is evidence that a reaction has produced a substance that does not remain dissolved. To calculate how much reactant or product is involved, combine two ideas: concentration and volume give the amount of dissolved substance, and a balanced chemical equation gives the reacting amount ratio. This lesson reviews those ideas and applies them in a calculation.

What you will learn

Explain how concentration and volume can be used to find the amount of a dissolved substance.
Use a balanced chemical equation to relate amounts of substances in a reaction.
Solve a solution-stoichiometry problem and report the answer with suitable units and significant digits.

1. Bridge: amount, volume, and concentration

A solution is a uniform mixture. The substance dissolved in it is the solute. Concentration describes how much solute is present in a given volume of solution. In this lesson, concentration is given in moles per litre, written as mol/L. The symbol

n

means amount in moles,

c

means concentration, and

V

means volume.

The relationship

n=cV

gives the amount of dissolved solute when concentration and volume are known. It works when the volume is in litres. If a problem gives millilitres, convert to litres before using the relationship. For example,

25.0\ \mathrm{mL}=0.0250\ \mathrm{L}

. Keep units in each line so you can see whether they cancel correctly.

A mole is a counting unit for particles. You do not need to count individual ions in a solution. Instead, the stated concentration and volume let you calculate the amount of solute in moles. This is the starting point for solution stoichiometry.

n=cV

Use the volume of the solution, not the volume of solute added to make it.
Convert millilitres to litres before using $n=cV$ .
When concentration is in mol/L and volume is in L, the calculated amount is in moles.

2. From what you observe to the reaction ratio

Suppose two clear solutions are mixed and a solid forms. At the particle level, dissolved ions move among water particles. During the reaction, some ions join in a new combination and form particles of a solid. Other ions may remain dissolved. This particle model helps explain why a visible solid can appear.

A balanced chemical equation represents the reaction. Its coefficients show the mole ratio: the relative amounts of each substance that react or form. A coefficient is the number placed before a formula. For example, in

\mathrm{CaCl_2(aq)+Na_2CO_3(aq)\rightarrow CaCO_3(s)+2NaCl(aq)}

, one mole of calcium chloride reacts with one mole of sodium carbonate. One mole of calcium carbonate forms.

The state labels identify whether a substance is aqueous, meaning dissolved in water, or solid. In this example, calcium carbonate is the solid product. The balanced equation conserves each kind of atom. Its coefficients must not be changed to fit a calculation; they come from balancing the reaction.

A solution-stoichiometry problem usually follows this path: use concentration and volume to find moles of a known dissolved reactant, use the balanced equation's mole ratio to find moles of the substance asked for, then convert those moles to the requested unit if needed. If amounts of more than one reactant are given, identify the limiting reactant before calculating product. The limiting reactant is the reactant that runs out first. In the example below, the other reactant is stated to be in excess, so it does not limit the amount of product.

\frac{n(\mathrm{CaCO_3})}{n(\mathrm{CaCl_2})}=\frac{1}{1}

A balanced equation gives mole ratios through its coefficients.
Use the coefficient ratio for the substances named in the question.
An excess reactant is present in more than the amount needed to react with the limiting reactant.

3. A reliable calculation method

First, write the balanced reaction and identify the substance whose solution concentration and volume are known. Calculate its amount with

n=cV

. Next, use the coefficients in the equation to find the amount of the requested substance. This step is a mole-ratio conversion: it changes the amount of one substance into the amount of another using the balanced equation.

If the question asks for mass, multiply the amount in moles by the molar mass. Molar mass is the mass of one mole of a substance, usually expressed in g/mol. If the question asks for a concentration, volume, or another quantity, rearrange the relevant relationship only after finding the amount needed.

Track units through the work. In

n=cV

, the units mol/L and L combine to give mol. In a mass calculation, mol multiplied by g/mol gives g. Round the final answer to match the precision of the given measurements. Do not round intermediate values too early, because early rounding can affect the final result.

m=nM

Calculate moles from solution data before using the reaction ratio.
Use the balanced equation, not the subscripts in the formulas, to choose the mole ratio.
Keep extra digits during the calculation and round the final answer appropriately.

4. Units and reasonableness

A concentration written as

0.200\ \mathrm{mol/L}

means that each litre of solution contains

0.200

mol of solute. A smaller sample of that same solution contains a proportionally smaller amount. This is why volume must be included in the calculation.

Check that the answer has the requested kind of unit. A question asking for mass should end in grams, not moles. A question asking for amount should end in moles. Also check that the answer is reasonable: a small volume of a moderately concentrated solution should not produce an amount larger than the total amount of reactant calculated from that solution.

Significant digits communicate the precision of measured values. In the worked example, both volume and concentration have three significant digits, so the final mass is reported to three significant digits. Exact coefficients in a balanced equation do not limit the significant digits.

Check the final unit against what the question asks for.
Use the given measured values to decide how many significant digits to report.
Use a size check to catch unit-conversion or mole-ratio errors.

Worked example

Finding the mass of a precipitate

A

25.0\ \mathrm{mL}

sample of

0.200\ \mathrm{mol/L}

calcium chloride solution reacts with excess sodium carbonate solution. What mass of calcium carbonate forms?

Write the balanced reaction
Calcium carbonate is the solid product. The balanced equation shows that calcium chloride and calcium carbonate have a 1:1 mole ratio.
$\mathrm{CaCl_2(aq)+Na_2CO_3(aq)\rightarrow CaCO_3(s)+2NaCl(aq)}$
Convert the volume
The concentration is in moles per litre, so convert the given volume from millilitres to litres.
$25.0\ \mathrm{mL}\times\frac{1\ \mathrm{L}}{1000\ \mathrm{mL}}=0.0250\ \mathrm{L}$
Find calcium chloride amount
Multiply concentration by volume. The litre units cancel, leaving moles of calcium chloride.
$n(\mathrm{CaCl_2})=(0.200\ \mathrm{mol/L})(0.0250\ \mathrm{L})=0.00500\ \mathrm{mol}$
Use the mole ratio
The equation has one calcium chloride for each calcium carbonate. Therefore, the amount of calcium carbonate formed is also $0.00500\ \mathrm{mol}$ .
$0.00500\ \mathrm{mol\ CaCl_2}\times\frac{1\ \mathrm{mol\ CaCO_3}}{1\ \mathrm{mol\ CaCl_2}}=0.00500\ \mathrm{mol\ CaCO_3}$
Convert amount to mass
The molar mass of calcium carbonate is about $100.09\ \mathrm{g/mol}$ . Multiply by the amount, then report three significant digits.
$m=(0.00500\ \mathrm{mol})(100.09\ \mathrm{g/mol})=0.500\ \mathrm{g}$

Answer: The reaction forms

0.500\ \mathrm{g}

of calcium carbonate.

Check: The sample contains only

0.00500\ \mathrm{mol}

of calcium chloride, so a mass of about half a gram of calcium carbonate is reasonable. The final unit is grams, as requested.

Common mistakes and how to avoid them

Using millilitres directly in

n=cV

when concentration is in mol/L.

Correction: Convert the volume to litres first. Otherwise, the units do not give an amount in moles.

Using a formula subscript as a mole ratio.

Correction: Use the coefficients in the balanced equation. Subscripts show how many atoms are in a formula, not how many moles react.

Changing the balanced equation's coefficients to match the amounts in the question.

Correction: Keep the balanced equation fixed. Use its coefficient ratio to relate the calculated amounts.

Reporting moles when the question asks for mass.

Correction: After finding moles, use the molar mass to convert to grams.

Rounding each intermediate answer heavily.

Correction: Keep extra digits during the calculation, then round the final answer to the appropriate significant digits.

Lesson summary

Use $n=cV$ to calculate moles from concentration and volume, with volume in litres.
Use coefficients from a balanced equation to convert between amounts of reacting substances.
Convert moles to the requested quantity, keep units visible, and round the final answer appropriately.

Check your understanding

Question 1

What volume in litres should be used for a

40.0\ \mathrm{mL}

solution in

n=cV

?

$0.0400\ \mathrm{L}$
$0.400\ \mathrm{L}$
$4.00\ \mathrm{L}$
$40000\ \mathrm{L}$

Show answer and explanation

0.0400\ \mathrm{L}

There are

1000\ \mathrm{mL}

in one litre, so

40.0\ \mathrm{mL}=0.0400\ \mathrm{L}

.

Question 2

A

10.0\ \mathrm{mL}

sample of

0.150\ \mathrm{mol/L}

solution contains how many moles of solute?

$0.00150\ \mathrm{mol}$
$0.0150\ \mathrm{mol}$
$0.150\ \mathrm{mol}$
$1.50\ \mathrm{mol}$

Show answer and explanation

0.00150\ \mathrm{mol}

Convert

10.0\ \mathrm{mL}

to

0.0100\ \mathrm{L}

, then calculate

(0.150\ \mathrm{mol/L})(0.0100\ \mathrm{L})=0.00150\ \mathrm{mol}

.

Question 3

In the balanced reaction

\mathrm{2HCl(aq)+Mg(s)\rightarrow MgCl_2(aq)+H_2(g)}

, what is the mole ratio of hydrochloric acid to hydrogen gas?

1:1
2:1
1:2
2:3

Show answer and explanation

2:1

The coefficients show that two moles of hydrochloric acid are used for each mole of hydrogen gas formed.

Key terms

Solution: A uniform mixture in which one or more substances are dissolved.
Solute: A substance dissolved in a solution.
Concentration: The amount of solute in a stated volume of solution.
Mole ratio: A ratio between amounts of substances taken from the coefficients of a balanced chemical equation.
Molar mass: The mass of one mole of a substance, commonly expressed in grams per mole.
Limiting reactant: The reactant that runs out first and sets the maximum amount of product that can form.

Continue through SCH3U

View the complete SCH3U Ontario Grade 11 Chemistry curriculum and lessons

Tutor support for E2.6 · Solve solution-stoichiometry problems

Working on E2.6 · Solve solution-stoichiometry problems in SCH3U? Bring the question and your own attempt. In a live session, you can write the chemical relationship, check conservation and units, then trace each calculation. Ask about the first step that does not make sense, then try a similar question yourself.

Explore chemistry tutoring · Request an eligible free 30-minute introduction. Your request is matched by subject and language; a tutor must accept before a time is confirmed.

برای مبحث E2.6 · حل مسائل استوکیومتری محلول در درس SCH3U، سؤال و تلاش خود را همراه بیاورید. در جلسه می‌توانید رابطهٔ شیمیایی را بنویسید، پایستگی و یکاها را بررسی کنید و هر محاسبه را دنبال کنید. راهنمای معلم شیمی را ببینید یا جلسهٔ معارفهٔ رایگان ۳۰ دقیقه‌ای درخواست کنید. زمان جلسه پس از پذیرش معلم قطعی می‌شود.

About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 11 Chemistry (SCH3U), expectation E2.6. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

Official curriculum reference

Report a correction or ask a question