How to Predict the Salt Formed by a Metal and an Acid

A strip of magnesium is dropped into dilute hydrochloric acid. Bubbles appear, and the magnesium slowly disappears. But what salt is left in the liquid: magnesium chloride, MgCl, or MgCl₂?

The answer is magnesium chloride, MgCl₂. To predict the salt produced by a metal and an acid, combine the positive ion formed by the metal with the negative ion supplied by the acid. Then balance their charges so the salt is electrically neutral.

This method works for the standard reactions between reactive metals and dilute, non-oxidising acids. The important part is knowing which ions come from each reactant. Once you can identify those, predicting the salt becomes much easier.

01What actually happens when a metal reacts with an acid?

Start with the magnesium and hydrochloric acid reaction.

An illustrative school investigation uses a small piece of magnesium ribbon and dilute hydrochloric acid in a flask. Initially, the experiment might simply use an open test tube, allowing students to observe the reaction.

The prediction is that magnesium will react with the acid, producing hydrogen gas and a dissolved salt.

What should you observe?

Three observations are expected:

• The magnesium ribbon gradually becomes smaller as it reacts.
• Bubbles form, and gas can be collected if a gas syringe is attached.
• The remaining solution is colourless.

However, observations and explanations are not the same thing.

The key point is that you cannot see magnesium chloride as a solid sitting in the solution. It dissolves.

The reaction produces magnesium ions and chloride ions, which remain dispersed throughout the water.

Why was the experimental design improved?

An open test tube allows gas to escape. You can see bubbles, but you cannot reliably measure how much gas has formed.

A gas syringe improves the experiment. The flask is sealed with a bung, and a delivery tube directs the gas into the syringe. As gas enters, the syringe plunger moves outwards.

This allows the volume of gas to be measured.

Even the improved apparatus has a limitation: measuring gas volume does not directly identify the dissolved salt. The salt’s identity must be inferred from the reacting ions and their charges.

From observations to particles

Magnesium atoms lose electrons to form magnesium ions:

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

Hydrogen ions from the acid gain those electrons:

\[ \ce{2H+ + 2e- -> H2} \]

Combining the two half-equations gives the net ionic equation:

\[ \ce{Mg + 2H+ -> Mg^2+ + H2} \]

Notice that chloride ions do not appear in this net ionic equation.

They are spectator ions: they remain in the solution while magnesium reacts with hydrogen ions.

The complete chemical equation is:

\[ \ce{Mg + 2HCl -> MgCl2 + H2} \]

This explains where every part of the products comes from:

• Magnesium supplies the \(\mathrm{Mg}^{2+}\) ions.
• Hydrochloric acid supplies the \(\mathrm{Cl}^{-}\) ions.
• Hydrogen ions form hydrogen gas.
• Magnesium ions and chloride ions make up the salt, magnesium chloride.

This last distinction is important. In aqueous solution, magnesium chloride exists as hydrated ions, not as separate MgCl₂ molecules. The formula MgCl₂ represents the simplest whole-number ratio of ions in the ionic compound.

02The method: how to predict the salt formula

You only need three pieces of information:

1. The metal’s positive ion.
2. The acid’s negative ion.
3. The charge balance needed to make the salt neutral.

Let’s build each part.

Step 1: Identify the metal ion

The metal becomes a positive ion by losing electrons.

For many common metals, their charges are predictable.

Transition metals need a little more care because they can form ions with different charges.

For example, iron commonly forms \(\mathrm{Fe}^{2+}\) when reacting with dilute, non-oxidising acids. However, iron can also form \(\mathrm{Fe}^{3+}\) in other chemical reactions.

If a question gives you a Roman numeral, use it to identify the charge:

• Iron(II) means \(\mathrm{Fe}^{2+}\).
• Iron(III) means \(\mathrm{Fe}^{3+}\).

Never assume every metal has a charge of +2.

Step 2: Identify the negative ion from the acid

The acid supplies the negative ion, also called its anion.

For the common acids used in metal-acid reactions, learn these three:

The hydrogen ions are not part of the salt. They are involved in producing hydrogen gas in the standard dilute-acid reaction.

For example:

• Hydrochloric acid produces chloride salts.
• Sulfuric acid produces sulfate salts.
• Nitric acid supplies nitrate ions for nitrate salts.

Be careful not to confuse sulfate and sulfide. Sulfate is \(\mathrm{SO_{4}}^{2-}\), whereas sulfide is \(\mathrm{S}^{2-}\).

Step 3: Combine the ions so their charges cancel

This is the most important step.

An ionic compound must have an overall charge of zero. The total positive charge must equal the total negative charge.

Think of it as balancing two sides of an equation.

Consider magnesium chloride.

Magnesium forms \(\mathrm{Mg}^{2+}\), and hydrochloric acid supplies \(\mathrm{Cl}^{-}\).

One magnesium ion has a charge of +2. Each chloride ion has a charge of -1.

Therefore, you need two chloride ions for every magnesium ion:

\[ (+2)+(-1)+(-1)=0 \]

The formula is:

\[ \boxed{\ce{MgCl2}} \]

You can use the charge-crossing method to help with more complicated formulas. Take the magnitude of each ion’s charge and use it as the subscript of the other ion. Then simplify if possible.

Here are some useful combinations:

One more detail: when you need more than one polyatomic ion, put the entire ion in brackets.

For example, aluminium sulfate is \(\ce{Al2(SO4)3}\), not \(\ce{Al2SO43}\).

The brackets show that there are three complete sulfate ions.

03Worked examples: predicting the salt

Now put the method into practice.

04A useful experimental calculation: checking the predicted reaction

The salt formula comes from charge balance, but experimental measurements can help check whether the proposed reaction is consistent with the observations.

Consider an illustrative investigation using:

• 0.0600 g of magnesium.
• Excess dilute hydrochloric acid.
• A gas syringe to collect hydrogen.
• A temperature of 25.0 °C and a pressure of 100 kPa.

Suppose the illustrative gas volume is 61.0 mL.

Can the predicted reaction account for this volume?

The balanced equation is:

\[ \ce{Mg + 2HCl -> MgCl2 + H2} \]

The mole ratio between magnesium and hydrogen is 1:1.

Step 1: Calculate the amount of magnesium.

Using \(n=m/M\), where \(n\) is the amount in moles, \(m\) is mass in grams, and \(M\) is molar mass in grams per mole:

\[ n(\ce{Mg})=\frac{0.0600}{24.3} \]

\[ =0.00247\text{ mol} \]

Step 2: Use the mole ratio.

One mole of magnesium produces one mole of hydrogen.

Therefore:

\[ n(\ce{H2})=0.00247\text{ mol} \]

Step 3: Calculate the predicted gas volume.

At 25 °C and 100 kPa, use a molar gas volume of approximately 24.8 L mol⁻¹.

\[ V=nV_m \]

\[ V=0.00247\times24.8 \]

\[ V=0.0613\text{ L} \]

Converting to millilitres:

\[ \boxed{V=61.3\text{ mL}} \]

The predicted volume of 61.3 mL is close to the illustrative value of 61.0 mL.

This is consistent with the proposed reaction and its 1:1 magnesium-to-hydrogen mole ratio.

However, gas volume alone does not establish the identity of the dissolved salt. That still comes from identifying the ions and balancing their charges.

The calculation assumes complete reaction, excess acid, no gas loss, and the stated gas conditions.

05When does this method not work?

There is an important condition behind the familiar equation:

\[ \text{Metal + acid} \rightarrow \text{Salt + hydrogen} \]

The metal must actually react with the acid.

For the standard hydrogen-producing reaction, the metal must be sufficiently reactive, and the acid should be dilute and non-oxidising.

Case 1: The metal is not reactive enough

Consider copper and dilute hydrochloric acid.

A student might reason:

• Copper forms \(\mathrm{Cu}^{2+}\).
• Hydrochloric acid supplies \(\mathrm{Cl}^{-}\).
• Therefore, the salt must be \(\ce{CuCl2}\).

The formula \(\ce{CuCl2}\) is a valid neutral ionic formula. But that does not mean it will form in this reaction.

Copper is below hydrogen in the metal reactivity series. It does not react with dilute hydrochloric acid under ordinary conditions.

Therefore, no copper chloride is produced.

This exposes an important distinction:

Knowing the possible salt formula is not enough. You must first establish whether the reaction occurs.

For more detail on deciding which metals react with dilute acids, see Metal Reactivity with Dilute Acids: HSC Chemistry Guide.

Case 2: The acid is oxidising

There is another complication.

Nitric acid supplies nitrate ions, \(\ce{NO3-}\), so it can produce metal nitrate salts.

However, nitric acid is also an oxidising agent. Its reactions with metals do not generally follow the simple hydrogen-producing pattern.

For example, copper can react with nitric acid to form copper nitrate, even though copper does not react with dilute hydrochloric acid.

The gaseous products are nitrogen oxides rather than the hydrogen expected from a standard dilute, non-oxidising acid reaction. The exact products depend on the conditions.

Similarly, concentrated sulfuric acid can act as an oxidising agent, so you should not automatically predict hydrogen gas.

The important distinction is:

• The acid determines which anion may appear in the salt.
• The reaction conditions determine whether the metal reacts and what other products form.

For ordinary metal and dilute hydrochloric or sulfuric acid questions, the familiar salt-plus-hydrogen pattern is the appropriate starting point.

06Common mistakes to avoid

07Questions and solutions

Try these questions without looking back at the worked examples. They move from straightforward salt prediction to combining charge balance, reaction stoichiometry, experimental evidence, and redox reasoning.

Question 1

Calcium metal reacts with excess dilute hydrochloric acid.

Identify the salt produced, write its chemical formula, and provide the balanced chemical equation.

Solution 1

The salt is calcium chloride, \(\ce{CaCl2}\).

Calcium forms \(\mathrm{Ca}^{2+}\), while hydrochloric acid supplies \(\mathrm{Cl}^{-}\).

One calcium ion requires two chloride ions to produce a neutral compound:

\[ (+2)+2(-1)=0 \]

Therefore:

\[ \ce{CaCl2} \]

The unbalanced equation is:

\[ \ce{Ca + HCl -> CaCl2 + H2} \]

Two hydrochloric acid molecules are required to supply the two chloride ions and two hydrogen ions.

The balanced equation is:

\[ \boxed{\ce{Ca + 2HCl -> CaCl2 + H2}} \]

This produces one mole of calcium chloride for every mole of calcium that reacts.

Question 2

A student reacts 0.20 mol of aluminium with 0.50 mol of hydrochloric acid.

Assume the reaction goes to completion wherever possible and that the acid is dilute and in excess only if its calculated amount permits.

(a) Identify the salt and write the balanced chemical equation.

(b) Determine the limiting reactant.

(c) Calculate the amount, in moles, of salt produced.

(d) Calculate the amount of aluminium remaining after the reaction.

Solution 2

The salt is aluminium chloride, \(\ce{AlCl3}\), and hydrochloric acid is the limiting reactant.

Step 1: Predict the salt formula.

Aluminium forms \(\mathrm{Al}^{3+}\), while hydrochloric acid supplies \(\mathrm{Cl}^{-}\).

Three chloride ions are needed to balance one aluminium ion:

\[ (+3)+3(-1)=0 \]

Therefore:

\[ \ce{AlCl3} \]

Step 2: Balance the chemical equation.

\[ \boxed{\ce{2Al + 6HCl -> 2AlCl3 + 3H2}} \]

Step 3: Determine the limiting reactant.

From the balanced equation:

2 mol Al react with 6 mol HCl.

Therefore, 1 mol Al requires 3 mol HCl.

For 0.20 mol of aluminium, the required amount of hydrochloric acid is:

\[ n(\ce{HCl})=0.20\times3 \]

\[ =0.60\text{ mol} \]

Only 0.50 mol of hydrochloric acid is available.

Therefore, hydrochloric acid is the limiting reactant.

Step 4: Calculate the amount of aluminium chloride produced.

From the equation:

6 mol HCl produce 2 mol AlCl₃.

Therefore:

\[ n(\ce{AlCl3})=0.50\times\frac{2}{6} \]

\[ \boxed{n(\ce{AlCl3})=0.167\text{ mol}} \]

Step 5: Calculate the remaining aluminium.

The mole ratio between aluminium and aluminium chloride is 1:1.

Therefore, 0.167 mol of aluminium reacts.

The amount remaining is:

\[ n(\ce{Al})_{\text{remaining}} =0.20-0.167 \]

\[ \boxed{n(\ce{Al})_{\text{remaining}}=0.033\text{ mol}} \]

The important lesson is that predicting the salt formula and balancing the equation must happen before you can correctly perform limiting-reactant calculations.

Question 3

A student compares the reactions of zinc and copper with dilute hydrochloric acid.

Two equal-volume samples of hydrochloric acid have the same concentration and temperature. Equal-sized, cleaned pieces of zinc and copper are placed in separate reaction vessels.

After five minutes, the following illustrative data are recorded:

The gas produced in the zinc reaction is identified as hydrogen. The gas syringes have a measurement resolution of 1.0 mL.

(a) Predict the salt produced in the zinc reaction.

(b) Explain why copper does not produce copper chloride under these conditions.

(c) Explain why the difference in gas volumes does not mean that the zinc salt and copper salt have different formulas.

(d) What does the copper measurement actually establish, and what limitation does it have?

Solution 3

Zinc produces zinc chloride, \(\ce{ZnCl2}\), while copper shows no measurable reaction with dilute hydrochloric acid under the stated conditions.

(a) Zinc forms \(\mathrm{Zn}^{2+}\), and hydrochloric acid supplies \(\mathrm{Cl}^{-}\).

Two chloride ions are required for each zinc ion:

\[ \boxed{\ce{ZnCl2}} \]

The balanced equation is:

\[ \ce{Zn + 2HCl -> ZnCl2 + H2} \]

(b) Copper is less reactive than hydrogen.

It cannot displace hydrogen from dilute hydrochloric acid under ordinary conditions. Therefore, the predicted copper chloride formula does not become an actual reaction product.

(c) Gas volume and salt formula describe different aspects of the reaction.

The gas volume relates to how much hydrogen is produced. The salt formula depends on the charges of the metal ion and acid anion.

Zinc produces hydrogen and zinc chloride because it reacts. Copper does not produce a measurable amount of hydrogen because it does not react under these conditions.

It would be incorrect to use the relative gas volumes to determine the subscripts in the salt formula.

(d) The copper result establishes that less than 1.0 mL of gas was collected during the five-minute observation period.

It does not prove that absolutely no gas could ever form. The instrument has limited resolution, and the conclusion applies to the stated conditions and observation time.

Combined with the known reactivity of copper, the result is consistent with no reaction occurring.

Question 4

Iron reacts with excess dilute sulfuric acid.

A student observes gas bubbles and a pale green solution. They propose the following salt formula:

\[ \ce{Fe(SO4)2} \]

Their reasoning is that sulfuric acid contains two hydrogen atoms, so two sulfate units must be needed in the salt.

(a) Explain why this reasoning is incorrect.

(b) Identify the correct salt formula, assuming iron forms \(\mathrm{Fe}^{2+}\) in this reaction.

(c) Write the balanced molecular equation and the net ionic equation.

(d) Identify the spectator ion and explain which parts of the proposed reaction are observations and which are inferences.

Solution 4

The correct salt is iron(II) sulfate, \(\ce{FeSO4}\), not \(\ce{Fe(SO4)2}\).

(a) The student’s mistake is confusing the number of hydrogen atoms in the acid with the charge of the salt’s ions.

Sulfuric acid supplies sulfate ions, \(\mathrm{SO_{4}}^{2-}\), in the normal sulfate salt.

The iron ion is specified as \(\mathrm{Fe}^{2+}\).

The charges are already equal and opposite:

\[ (+2)+(-2)=0 \]

One iron ion therefore combines with one sulfate ion.

(b) The correct salt formula is:

\[ \boxed{\ce{FeSO4}} \]

(c) The balanced molecular equation is:

\[ \boxed{\ce{Fe + H2SO4 -> FeSO4 + H2}} \]

The net ionic equation is:

\[ \boxed{\ce{Fe + 2H+ -> Fe^2+ + H2}} \]

The sulfate ion does not undergo a chemical change in this standard dilute-acid reaction.

(d) Sulfate ions, \(\mathrm{SO_{4}}^{2-}\), are the spectator ions.

The observations are the gas bubbles, the dissolving iron, and the pale green appearance of the solution.

The inferences are that hydrogen gas is produced, iron forms \(\mathrm{Fe}^{2+}\), and sulfate ions remain in solution.

The pale green colour is consistent with iron(II) ions, but colour alone is not a complete identification of the dissolved particles.

The key distinction is that observations provide evidence, while chemical equations describe the particle-level explanation of that evidence.

Question 5

A pure, previously unidentified metal, M, reacts completely with excess dilute hydrochloric acid.

The following information is supplied:

• Amount of metal used: 0.0200 mol.
• Volume of hydrogen collected: 0.744 L.
• Temperature: 25.0 °C.
• Pressure: 100 kPa.
• Molar gas volume under these conditions: 24.8 L mol⁻¹.

Assume that:

• The metal forms only one type of positive ion during the reaction.
• Hydrogen ions are the only species being reduced.
• All the metal reacts, and no hydrogen is lost.
• The gas measurement is accurate.

The same metal is then reacted with excess dilute sulfuric acid. Assume it forms the same positive ion under these conditions.

(a) Calculate the amount of hydrogen produced.

(b) Use the electron transfer to determine the charge on the metal ion.

(c) Predict the formula of the salt formed with sulfuric acid.

(d) Write the balanced equation for the reaction with sulfuric acid.

(e) A student argues that the salt should be \(\ce{M2SO4}\) because sulfuric acid contains two hydrogen atoms. Explain precisely why this prediction fails.

Solution 5

The metal forms a 3+ ion, and its sulfate salt has the formula \(\ce{M2(SO4)3}\).

The important step is using the hydrogen measurement to determine how many electrons each metal atom loses. We can then use that charge to predict the sulfate formula.

(a) Calculate the amount of hydrogen.

Use:

\[ n=\frac{V}{V_m} \]

Where \(n\) is the amount of gas in moles, \(V\) is the gas volume in litres, and \(V_m\) is the molar gas volume in litres per mole.

Substitute:

\[ n(\ce{H2})=\frac{0.744}{24.8} \]

\[ \boxed{n(\ce{H2})=0.0300\text{ mol}} \]

(b) Determine the charge on the metal ion.

Each hydrogen molecule requires two hydrogen ions to gain two electrons:

\[ \ce{2H+ + 2e- -> H2} \]

Therefore, producing 0.0300 mol of hydrogen requires:

\[ n(e^-)=2\times0.0300 \]

\[ =0.0600\text{ mol electrons} \]

We know that 0.0200 mol of metal atoms supplied these electrons.

Therefore, the number of electrons lost per mole of metal atoms is:

\[ \frac{0.0600}{0.0200}=3.00 \]

Each metal atom loses three electrons.

The oxidation half-equation is:

\[ \ce{M -> M^3+ + 3e-} \]

Therefore, the metal ion is:

\[ \boxed{\mathrm{M}^{3+}} \]

(c) Predict the sulfate salt formula.

Sulfuric acid supplies sulfate ions:

\[ \mathrm{SO_{4}}^{2-} \]

The metal ion has a charge of +3, and the sulfate ion has a charge of -2.

We need to balance the charges.

Two metal ions provide:

\[ 2\times(+3)=+6 \]

Three sulfate ions provide:

\[ 3\times(-2)=-6 \]

The total charge is zero.

Therefore, the salt formula is:

\[ \boxed{\ce{M2(SO4)3}} \]

(d) Write the balanced chemical equation.

Start with:

\[ \ce{M + H2SO4 -> M2(SO4)3 + H2} \]

The salt requires two metal atoms and three sulfate ions.

Therefore, use two metal atoms and three sulfuric acid molecules:

\[ \ce{2M + 3H2SO4 -> M2(SO4)3 + H2} \]

There are six hydrogen atoms on the left, so three hydrogen molecules must form on the right.

The final balanced equation is:

\[ \boxed{\ce{2M + 3H2SO4 -> M2(SO4)3 + 3H2}} \]

(e) Explain why the student’s prediction is incorrect.

The student has confused the number of hydrogen atoms in sulfuric acid with the charge balance required in the salt.

The two hydrogen atoms in sulfuric acid are involved in producing hydrogen gas. The remaining sulfate ion has a charge of -2.

The metal ion has a charge of +3, as established by the first experiment.

If we used \(\ce{M2SO4}\), the total charge would be:

\[ 2(+3)+(-2)=+4 \]

The compound would have an overall positive charge, so it could not represent a neutral ionic salt.

We need three sulfate ions to balance the total +6 charge from two metal ions.

This question also demonstrates why experimental measurements can reveal information that isn’t obvious from the reactant names alone. The hydrogen volume allowed us to determine the metal’s oxidation state, which then determined the salt formula.

08What to remember

When predicting the salt from a metal and an acid, keep the roles of the reactants separate:

• The metal supplies the positive ion.
• The acid supplies the negative ion.
• The ion charges determine the salt formula.
• The balanced equation determines the mole ratios.
• The reactivity series and acid conditions determine whether the reaction actually occurs.

Once you can predict the salt correctly, the next useful step is reaction stoichiometry: using the balanced equation to calculate limiting reactants, product amounts, and gas volumes. Always establish the correct products before starting those calculations.