How to Solve Stoichiometry Using a Balanced Ionic Equation

Suppose you dissolve 0.800 g of sodium hydroxide and add it to a solution containing excess magnesium chloride. A white solid forms. You want to calculate the mass of that solid before collecting it.

The balanced net ionic equation is:

\[ \ce{Mg^2+(aq) + 2OH^-(aq) -> Mg(OH)2(s)} \]

The coefficient of hydroxide is 2. Does that mean 2 g of sodium hydroxide produces 1 g of magnesium hydroxide?

No. The coefficients represent mole ratios, not mass ratios.

To calculate the mass of a product from the mass of a reactant, follow this route:

\[ \boxed{\text{Mass} \rightarrow \text{Moles} \rightarrow \text{Moles} \rightarrow \text{Mass}} \]

More precisely, you convert the mass of your starting compound into moles, identify how many moles of the reacting ion it supplies, use the balanced ionic equation to calculate the moles of product, and finally convert those moles into mass.

The important skill is knowing how to move between the original compound and the ions that actually appear in the ionic equation.

01What information does a balanced ionic equation give you?

Consider the reaction between magnesium chloride and sodium hydroxide.

The balanced molecular equation is:

\[ \ce{MgCl2(aq) + 2NaOH(aq) -> Mg(OH)2(s) + 2NaCl(aq)} \]

When these substances dissolve, they separate into ions:

\[ \ce{MgCl2(aq) -> Mg^2+(aq) + 2Cl^-(aq)} \]

\[ \ce{NaOH(aq) -> Na^+(aq) + OH^-(aq)} \]

The magnesium ions and hydroxide ions form an insoluble solid, magnesium hydroxide.

Sodium and chloride ions remain dissolved. They are spectator ions because they do not participate in the actual chemical change.

Removing these spectator ions gives the net ionic equation:

\[ \ce{Mg^2+(aq) + 2OH^-(aq) -> Mg(OH)2(s)} \]

This equation tells us that:

• 1 mole of magnesium ions reacts with 2 moles of hydroxide ions.
• 1 mole of magnesium hydroxide forms for every 1 mole of magnesium ions consumed.
• 1 mole of magnesium hydroxide requires 2 moles of hydroxide ions.

These relationships are exactly what you need for stoichiometry.

The equation gives mole ratios, not compound ratios

There is another detail worth understanding.

The ionic equation contains \(\mathrm{Mg}^{2+}\), not \(\ce{MgCl2}\). However, if your question gives you the mass of magnesium chloride, you can still use it.

Why?

Because each formula unit of magnesium chloride supplies one magnesium ion:

\[ \ce{MgCl2 -> Mg^2+ + 2Cl^-} \]

Therefore:

\[ 1\text{ mol MgCl}_2 \rightarrow 1\text{ mol Mg}^{2+} \]

Likewise:

\[ 1\text{ mol NaOH} \rightarrow 1\text{ mol OH}^- \]

You must establish this relationship before using the coefficients in the ionic equation.

This becomes particularly important when the starting compound contains more than one of the reacting ions per formula unit.

For example:

\[ \ce{Al2(SO4)3 -> 2Al^3+ + 3SO4^2-} \]

One mole of aluminium sulfate supplies two moles of aluminium ions, not one.

If you need to revise how to construct balanced equations for acid reactions, see Balanced Equations for Acid Reactions in HSC Chemistry. For a closer look at ions that can donate or accept protons, see Amphiprotic Salts and Ionic Equations.

02The mass-to-moles-to-moles-to-mass method

There are four main calculation steps.

Step 1: Convert the known mass into moles

Use:

\[ n=\frac{m}{M} \]

Where:

• \(n\) is the amount of substance, measured in mol.
• \(m\) is the mass, measured in g.
• \(M\) is the molar mass, measured in \(\mathrm{g\,mol^{-1}}\).

Always use the molar mass of the compound actually weighed.

For example, if the question gives you the mass of sodium hydroxide, calculate moles using the molar mass of \(\ce{NaOH}\), not \(\mathrm{OH}^{-}\).

Step 2: Convert moles of the compound into moles of the reacting ion

Check the chemical formula.

If one formula unit supplies one reacting ion, the mole amounts are equal.

If one formula unit supplies two reacting ions, multiply by two.

For example:

Do not skip this step just because the compound and ion have similar-looking formulas.

Step 3: Use the ionic equation to calculate moles of product

Take the coefficients from the balanced net ionic equation.

For:

\[ \ce{Mg^2+ + 2OH^- -> Mg(OH)2} \]

The mole ratio is:

\[ 1:2:1 \]

If you know the amount of hydroxide ions, calculate magnesium hydroxide using:

\[ n(\ce{Mg(OH)2}) = n(\mathrm{OH}^{-})\times\frac{1}{2} \]

If you know the amount of magnesium ions instead:

\[ n(\ce{Mg(OH)2}) = n(\mathrm{Mg}^{2+})\times\frac{1}{1} \]

The coefficients determine which fraction you use.

Step 4: Convert moles of product into mass

Once you have calculated the amount of product, use:

\[ m=nM \]

Make sure you use the molar mass of the product, not the starting compound.

04The common mistake: Treating coefficients as mass ratios

Let’s return to the original question.

A student sees:

\[ \ce{Mg^2+ + 2OH^- -> Mg(OH)2} \]

They might reason:

• 2 moles of hydroxide produce 1 mole of magnesium hydroxide.
• Therefore, 2 g of hydroxide produce 1 g of magnesium hydroxide.

The first statement is correct. The second is not.

The coefficients describe amounts in moles, not masses in grams.

Consider the actual masses involved.

Two moles of sodium hydroxide have a mass of:

\[ 2\times40.00=80.00\ \mathrm{g} \]

One mole of magnesium hydroxide has a mass of:

\[ 58.32\ \mathrm{g} \]

Therefore, the actual theoretical mass relationship is:

\[ 80.00\ \mathrm{g\ NaOH} \rightarrow 58.32\ \mathrm{g\ Mg(OH)2} \]

Not 2 g to 1 g.

For the original 0.800 g sample, the product mass is 0.583 g, as calculated.

A useful rule is:

Never apply an equation coefficient directly to a mass. Convert to moles first.

05What if both reactants are given?

So far, we have assumed that magnesium chloride is in excess.

But HSC questions often give you the quantities of both reactants. You then need to determine which reactant runs out first.

This is called identifying the limiting reagent.

The limiting reagent determines the maximum amount of product that can form.

The other reactant is in excess.

A reliable method for finding the limiting reagent

1. Convert the mass of each starting compound into moles.
2. Convert those amounts into moles of the reacting ions.
3. Use the balanced ionic equation to determine which reactant runs out first.
4. Calculate the product using the limiting reagent.
5. Convert the product amount into mass.

One useful shortcut is to divide the available moles of each reacting ion by its coefficient in the balanced equation.

The smaller result identifies the limiting reactant.

For example, consider:

\[ \ce{Al^3+ + 3OH^- -> Al(OH)3(s)} \]

If you have:

• 0.00400 mol of \(\mathrm{Al}^{3+}\)
• 0.0100 mol of \(\mathrm{OH}^{-}\)

Compare:

\[ \frac{0.00400}{1}=0.00400 \]

and

\[ \frac{0.0100}{3}=0.00333 \]

The smaller value is 0.00333 mol.

Hydroxide is the limiting reactant, and 0.00333 mol of aluminium hydroxide can form.

This method avoids the mistake of simply choosing whichever reactant has fewer moles. The coefficients matter.

06Questions and solutions

The following questions progress from a straightforward mass calculation to more demanding applications involving limiting reagents, hydrated salts, purity, and multiple ionic equations.

Use these atomic masses where required:

Question 1

0.300 g of pure sodium hydroxide is dissolved in water and added to excess magnesium chloride solution.

The reaction is represented by:

\[ \ce{Mg^2+(aq) + 2OH^-(aq) -> Mg(OH)2(s)} \]

Calculate the theoretical mass of magnesium hydroxide produced.

Solution 1

The theoretical mass of magnesium hydroxide is 0.219 g.

Step 1: Calculate the moles of sodium hydroxide.

\[ \begin{aligned} n(\ce{NaOH}) &=\frac{m}{M}\\ &=\frac{0.300}{40.00}\\ &=0.00750\ \mathrm{mol} \end{aligned} \]

Step 2: Calculate the moles of hydroxide ions.

One mole of sodium hydroxide supplies one mole of hydroxide ions.

\[ n(\mathrm{OH}^{-})=0.00750\ \mathrm{mol} \]

Step 3: Apply the mole ratio.

From the balanced ionic equation:

\[ 2\text{ mol OH}^- \rightarrow 1\text{ mol Mg(OH)}_2 \]

Therefore:

\[ \begin{aligned} n(\ce{Mg(OH)2}) &=0.00750\times\frac{1}{2}\\ &=0.00375\ \mathrm{mol} \end{aligned} \]

Step 4: Calculate the mass of magnesium hydroxide.

\[ \begin{aligned} m&=nM\\ &=0.00375\times58.32\\ &=0.2187\ \mathrm{g} \end{aligned} \]

Therefore:

\[ \boxed{m(\ce{Mg(OH)2})=0.219\ \mathrm{g}} \]

The magnesium chloride is in excess, so all the available hydroxide ions can react. The amount of sodium hydroxide determines the maximum precipitate mass.

Question 2

A student reacts 1.06 g of sodium carbonate with 0.392 g of pure sulfuric acid.

The balanced net ionic equation is:

\[ \ce{CO3^2-(aq) + 2H^+(aq) -> CO2(g) + H2O(l)} \]

Assume both acidic protons from sulfuric acid are available for complete reaction.

Calculate:

a) Which reactant is limiting.

b) The theoretical mass of carbon dioxide produced.

Solution 2

Sulfuric acid is the limiting reactant, and 0.176 g of carbon dioxide is produced.

Step 1: Calculate the moles of sodium carbonate.

The molar mass is:

\[ M(\ce{Na2CO3})=105.99\ \mathrm{g\,mol^{-1}} \]

Therefore:

\[ \begin{aligned} n(\ce{Na2CO3}) &=\frac{1.06}{105.99}\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

Each formula unit of sodium carbonate supplies one carbonate ion:

\[ n(\mathrm{CO_{3}}^{2-})=0.0100\ \mathrm{mol} \]

Step 2: Calculate the moles of sulfuric acid.

The molar mass is:

\[ M(\ce{H2SO4})=98.08\ \mathrm{g\,mol^{-1}} \]

Therefore:

\[ \begin{aligned} n(\ce{H2SO4}) &=\frac{0.392}{98.08}\\ &=0.00400\ \mathrm{mol} \end{aligned} \]

Each mole of sulfuric acid supplies two moles of acidic hydrogen ions overall.

Therefore:

\[ \begin{aligned} n(\mathrm{H}^{+}) &=0.00400\times2\\ &=0.00800\ \mathrm{mol} \end{aligned} \]

Step 3: Identify the limiting reagent.

The ionic equation requires:

\[ 1\text{ mol CO}_3^{2-}:2\text{ mol H}^+ \]

To react with all 0.0100 mol of carbonate ions, we would need:

\[ 0.0100\times2=0.0200\ \mathrm{mol\ H^+} \]

Only 0.00800 mol of hydrogen ions are available.

Therefore, sulfuric acid is limiting.

Step 4: Calculate the moles of carbon dioxide.

The equation shows that two moles of hydrogen ions produce one mole of carbon dioxide.

\[ \begin{aligned} n(\ce{CO2}) &=0.00800\times\frac{1}{2}\\ &=0.00400\ \mathrm{mol} \end{aligned} \]

Step 5: Convert the carbon dioxide amount into mass.

\[ M(\ce{CO2})=12.01+2(16.00) \]

\[ M(\ce{CO2})=44.01\ \mathrm{g\,mol^{-1}} \]

Therefore:

\[ \begin{aligned} m&=nM\\ &=0.00400\times44.01\\ &=0.17604\ \mathrm{g} \end{aligned} \]

The final answer is:

\[ \boxed{m(\ce{CO2})=0.176\ \mathrm{g}} \]

An important detail is that the ionic equation contains hydrogen ions, not sulfuric acid molecules. You must account for the two acidic hydrogen ions supplied by each sulfuric acid formula unit before applying the coefficient ratio.

Question 3

A student dissolves 0.500 g of magnesium chloride hexahydrate, \(\ce{MgCl2.6H2O}\), and mixes it with 0.300 g of sodium hydroxide.

The balanced net ionic equation is:

\[ \ce{Mg^2+(aq) + 2OH^-(aq) -> Mg(OH)2(s)} \]

Calculate:

a) The limiting reagent.

b) The theoretical mass of magnesium hydroxide produced.

Solution 3

Magnesium chloride hexahydrate is the limiting reagent, and 0.143 g of magnesium hydroxide is produced.

Step 1: Calculate the molar mass of magnesium chloride hexahydrate.

The six water molecules must be included in the molar mass.

\[ \begin{aligned} M(\ce{MgCl2.6H2O}) &=24.31+2(35.45)\\ &\quad+6(18.016)\\ &=203.30\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Step 2: Calculate the moles of magnesium chloride hexahydrate.

\[ \begin{aligned} n(\ce{MgCl2.6H2O}) &=\frac{0.500}{203.30}\\ &=0.00246\ \mathrm{mol} \end{aligned} \]

Each formula unit contains one magnesium ion.

Therefore:

\[ n(\mathrm{Mg}^{2+})=0.00246\ \mathrm{mol} \]

Step 3: Calculate the moles of hydroxide ions.

\[ \begin{aligned} n(\ce{NaOH}) &=\frac{0.300}{40.00}\\ &=0.00750\ \mathrm{mol} \end{aligned} \]

Since one mole of sodium hydroxide supplies one mole of hydroxide ions:

\[ n(\mathrm{OH}^{-})=0.00750\ \mathrm{mol} \]

Step 4: Identify the limiting reagent.

The reaction requires two moles of hydroxide for every mole of magnesium ions.

To react with all 0.00246 mol of magnesium ions, the required hydroxide amount is:

\[ \begin{aligned} n(\mathrm{OH}^{-})_{\text{required}} &=0.00246\times2\\ &=0.00492\ \mathrm{mol} \end{aligned} \]

We have 0.00750 mol of hydroxide ions available.

Therefore, hydroxide is in excess and magnesium ions are limiting.

Step 5: Calculate the moles of magnesium hydroxide.

The mole ratio between magnesium ions and magnesium hydroxide is 1:1.

\[ n(\ce{Mg(OH)2})=0.00246\ \mathrm{mol} \]

Step 6: Calculate the mass of magnesium hydroxide.

\[ \begin{aligned} m&=nM\\ &=0.00246\times58.32\\ &=0.143\ \mathrm{g} \end{aligned} \]

Therefore:

\[ \boxed{m(\ce{Mg(OH)2})=0.143\ \mathrm{g}} \]

The six water molecules in the hydrated salt contribute to its starting mass, but they do not provide additional magnesium ions. Ignoring the water of crystallisation would overestimate the amount of magnesium available.

Question 4

A laboratory requires 0.500 g of dry magnesium hydroxide.

The available sodium hydroxide sample is only 90.0% pure by mass. Magnesium chloride is available in excess.

Calculate the minimum mass of the impure sodium hydroxide sample required to produce the target mass of magnesium hydroxide.

The balanced net ionic equation is:

\[ \ce{Mg^2+ + 2OH^- -> Mg(OH)2} \]

Solution 4

The minimum mass of the impure sodium hydroxide sample required is 0.762 g.

This time, we work backwards from the desired product mass.

Step 1: Calculate the moles of magnesium hydroxide required.

The molar mass is:

\[ M(\ce{Mg(OH)2})=58.32\ \mathrm{g\,mol^{-1}} \]

Therefore:

\[ \begin{aligned} n(\ce{Mg(OH)2}) &=\frac{0.500}{58.32}\\ &=0.00857\ \mathrm{mol} \end{aligned} \]

Step 2: Calculate the moles of hydroxide ions required.

The ionic equation shows that two moles of hydroxide ions produce one mole of magnesium hydroxide.

Therefore:

\[ \begin{aligned} n(\mathrm{OH}^{-}) &=0.00857\times2\\ &=0.0171\ \mathrm{mol} \end{aligned} \]

Step 3: Calculate the mass of pure sodium hydroxide required.

One mole of sodium hydroxide supplies one mole of hydroxide ions.

Therefore:

\[ n(\ce{NaOH})=0.0171\ \mathrm{mol} \]

Using the molar mass of sodium hydroxide:

\[ \begin{aligned} m&=nM\\ &=0.0171\times40.00\\ &=0.686\ \mathrm{g} \end{aligned} \]

This is the mass of pure sodium hydroxide required.

Step 4: Account for the sample purity.

The sample is 90.0% pure, meaning 90.0% of its total mass is sodium hydroxide.

Therefore:

\[ \text{Mass of sample} = \frac{\text{Mass of pure NaOH}}{0.900} \]

Substitute:

\[ \begin{aligned} m_{\text{sample}} &=\frac{0.6857}{0.900}\\ &=0.762\ \mathrm{g} \end{aligned} \]

Therefore:

\[ \boxed{m_{\text{impure NaOH}}=0.762\ \mathrm{g}} \]

The important distinction is that 0.686 g is the mass of pure sodium hydroxide, not the mass of the entire impure sample.

Question 5

A 1.50 g sample contains a mixture of three anhydrous salts:

• Magnesium chloride, \(\ce{MgCl2}\)
• Magnesium sulfate, \(\ce{MgSO4}\)
• Sodium chloride, \(\ce{NaCl}\)

The sample is dissolved in water.

An excess of sodium hydroxide is added to the solution. All magnesium ions precipitate as magnesium hydroxide, which is filtered, washed, dried, and weighed.

The mass of dry magnesium hydroxide recovered is 0.583 g.

The combined filtrate and washings are then treated with excess barium chloride. All sulfate ions precipitate as barium sulfate, which is also filtered, washed, dried, and weighed.

The mass of dry barium sulfate recovered is 0.467 g.

Calculate the mass percentage of each original salt in the 1.50 g sample.

Assume:

• All three original salts dissolve completely.
• All magnesium ions precipitate in the first reaction.
• All sulfate ions precipitate in the second reaction.
• Sodium chloride remains dissolved throughout.
• Both precipitates are pure, dry, and recovered without loss.
• No other reactions occur.

The relevant net ionic equations are:

\[ \ce{Mg^2+ + 2OH^- -> Mg(OH)2(s)} \]

\[ \ce{Ba^2+ + SO4^2- -> BaSO4(s)} \]

Solution 5

The original mixture contains:

• 50.8% magnesium chloride
• 16.0% magnesium sulfate
• 33.2% sodium chloride

The difficult part is that the first precipitate does not tell us how much magnesium chloride was present by itself.

Both magnesium chloride and magnesium sulfate supply magnesium ions.

We need to use the second precipitation reaction to separate their contributions.

Step 1: Calculate the total moles of magnesium ions in the original mixture.

The first reaction produces magnesium hydroxide:

\[ \ce{Mg^2+ + 2OH^- -> Mg(OH)2} \]

The molar mass of magnesium hydroxide is:

\[ M(\ce{Mg(OH)2})=58.32\ \mathrm{g\,mol^{-1}} \]

The recovered precipitate has a mass of 0.583 g.

Therefore:

\[ \begin{aligned} n(\ce{Mg(OH)2}) &=\frac{0.583}{58.32}\\ &=0.0100\ \mathrm{mol} \end{aligned} \]

The mole ratio between magnesium ions and magnesium hydroxide is 1:1.

Therefore:

\[ n(\mathrm{Mg}^{2+})_{\text{total}} =0.0100\ \mathrm{mol} \]

This represents magnesium ions from both original magnesium salts combined.

Step 2: Calculate the moles of sulfate ions.

The second reaction produces barium sulfate:

\[ \ce{Ba^2+ + SO4^2- -> BaSO4(s)} \]

The molar mass of barium sulfate is:

\[ \begin{aligned} M(\ce{BaSO4}) &=137.33+32.06+4(16.00)\\ &=233.39\ \mathrm{g\,mol^{-1}} \end{aligned} \]

The recovered precipitate has a mass of 0.467 g.

Therefore:

\[ \begin{aligned} n(\ce{BaSO4}) &=\frac{0.467}{233.39}\\ &=0.00200\ \mathrm{mol} \end{aligned} \]

The mole ratio between sulfate ions and barium sulfate is 1:1.

Therefore:

\[ n(\mathrm{SO_{4}}^{2-})=0.00200\ \mathrm{mol} \]

Step 3: Calculate the moles and mass of magnesium sulfate.

Each formula unit of magnesium sulfate supplies one magnesium ion and one sulfate ion.

Therefore:

\[ n(\ce{MgSO4})=0.00200\ \mathrm{mol} \]

The molar mass of magnesium sulfate is:

\[ \begin{aligned} M(\ce{MgSO4}) &=24.31+32.06+4(16.00)\\ &=120.37\ \mathrm{g\,mol^{-1}} \end{aligned} \]

Calculate its mass:

\[ \begin{aligned} m(\ce{MgSO4}) &=0.00200\times120.37\\ &=0.241\ \mathrm{g} \end{aligned} \]

Step 4: Calculate the moles and mass of magnesium chloride.

From Step 1, the total amount of magnesium ions was:

\[ 0.0100\ \mathrm{mol} \]

From Step 3, magnesium sulfate supplied:

\[ 0.00200\ \mathrm{mol\ Mg^{2+}} \]

The remaining magnesium ions must have come from magnesium chloride.

Therefore:

\[ \begin{aligned} n(\ce{MgCl2}) &=0.0100-0.00200\\ &=0.00800\ \mathrm{mol} \end{aligned} \]

The molar mass of magnesium chloride is:

\[ M(\ce{MgCl2})=95.21\ \mathrm{g\,mol^{-1}} \]

Calculate its mass:

\[ \begin{aligned} m(\ce{MgCl2}) &=0.00800\times95.21\\ &=0.762\ \mathrm{g} \end{aligned} \]

Step 5: Calculate the mass of sodium chloride.

We know the total mass of the original sample:

\[ m_{\text{total}}=1.50\ \mathrm{g} \]

The mass of sodium chloride is the difference between the original sample mass and the masses of the two magnesium salts.

\[ \begin{aligned} m(\ce{NaCl}) &=1.50-0.241-0.762\\ &=0.498\ \mathrm{g} \end{aligned} \]

Step 6: Calculate the mass percentage of each salt.

Use:

\[ \text{Mass percentage} = \frac{\text{Mass of component}} {\text{Total sample mass}}\times100 \]

For magnesium chloride:

\[ \begin{aligned} \%\,\ce{MgCl2} &=\frac{0.762}{1.50}\times100\\ &=50.8\% \end{aligned} \]

For magnesium sulfate:

\[ \begin{aligned} \%\,\ce{MgSO4} &=\frac{0.241}{1.50}\times100\\ &=16.1\% \end{aligned} \]

For sodium chloride:

\[ \begin{aligned} \%\,\ce{NaCl} &=\frac{0.498}{1.50}\times100\\ &=33.2\% \end{aligned} \]

The final composition is approximately:

The slight difference from 100% is due to rounding.

The tempting approach would be to calculate the magnesium chloride mass directly from the first magnesium hydroxide precipitate.

That would incorrectly assume that every magnesium ion came from magnesium chloride.

The first ionic equation only tells us the total amount of magnesium ions. It cannot identify which original salt supplied them.

The second ionic equation provides the missing information by measuring the sulfate ions separately. Once we know how much magnesium sulfate was present, we can calculate the remaining magnesium chloride and finally use mass conservation to find the sodium chloride.

This is the important conceptual lesson: a balanced ionic equation tells you the mole relationship between reacting species, but you must consider where those species came from.

07Where to go next

Once you can calculate a theoretical product mass from a balanced ionic equation, the next step is interpreting experimental results.

A real experiment might produce less solid than predicted because some precipitate is lost during filtration, the reaction is incomplete, or the starting reactant is impure.

You then need to distinguish theoretical yield from actual yield and calculate percentage yield where appropriate.

Before doing that, make sure your stoichiometric calculation is correct: identify the reacting ions, account for the number of ions supplied by each formula unit, apply the balanced equation’s mole ratios, and only then calculate the final mass.