Metals with dilute acids

Drop a strip of zinc into dilute hydrochloric acid and the liquid fizzes at once, bubbles of gas streaming off the metal surface. Drop a strip of copper into the same acid and nothing happens at all — the copper just sits there, shiny and unreacted. The difference is not the acid; it is the metal, and specifically where that metal sits relative to hydrogen in the activity series. This article builds on the basic idea that reactive metals liberate hydrogen from dilute acids, adds the quantitative side — how much hydrogen gas a given mass of metal actually produces — and sharpens the reasoning behind the two big exceptions every Class 9 student must be able to explain: why metals below hydrogen refuse to react, and why dilute nitric acid, despite being a perfectly ordinary acid, never yields hydrogen gas.

The core insight

A metal liberates hydrogen from a dilute acid only when it lies above hydrogen in the activity series, following Metal + Dilute acid to Salt + Hydrogen, and the amount of gas produced follows directly from the reaction's mole ratio.

The core idea

Whether a metal reacts with a dilute acid is decided entirely by one comparison: is the metal more reactive than hydrogen itself? The activity series ranks metals (and hydrogen) by how readily they give up electrons to form positive ions. A metal above hydrogen can push hydrogen ions out of solution and take their place; a metal below hydrogen cannot. For the four metals prescribed in this syllabus — magnesium, aluminium, zinc, and iron — the general pattern is:

\text{Metal} + \text{Dilute acid} \rightarrow \text{Salt} + H_2\uparrow

where: the metal supplies electrons to the $H^+$ ions from the acid, reducing them to neutral $H_2$ gas, while the metal itself is oxidised into the corresponding metal ion that ends up in the salt. This single displacement reaction is why the test tube fizzes: each bubble is a molecule of $H_2$ escaping the liquid. Metals below hydrogen — copper, silver, gold — simply cannot supply electrons faster than hydrogen already holds onto them, so $H^+$ ions in the acid stay put and no gas forms. Dilute nitric acid is the odd one out among the common acids: it does contain $H^+$ ions, but it is such a strong oxidising agent that it grabs the newly-formed (nascent) hydrogen and oxidises it straight to water before it can escape as gas, while the acid itself is reduced to nitrogen oxides. So a reaction happens with dilute $HNO_3$ , but hydrogen is never the product.

Figure

A bar showing the activity series from most reactive on the left to least reactive on the right with hydrogen marked as the dividing line

Why it works

Beyond just naming which metals react, it helps to see the reasoning laid out in order. The dividing line is electron transfer, not acid strength. A metal reacts with a dilute acid only if it loses electrons more readily than hydrogen does — that is exactly what "above hydrogen in the activity series" means. So the check is always the same: locate the metal on the series, see whether it sits above or below hydrogen, and predict the outcome from that position alone. Salts follow the metal's valency. Once a metal is confirmed to react, the salt formed pairs the metal ion with the acid's anion in whatever ratio balances charge — magnesium (valency 2) gives $MgSO_4$ , aluminium (valency 3) gives $Al_2(SO_4)_3$ , and so on. Getting this ratio wrong is one of the most common marks lost in this topic. Nitric acid is excluded on oxidising grounds, not on activity-series grounds. Even a highly reactive metal like magnesium will react with dilute $HNO_3$ , but the acid's strong oxidising power intercepts the hydrogen before it can be released as gas, converting it to water while the acid is reduced to nitrogen oxides. The metal being reactive enough is necessary but not sufficient — the acid must also be non-oxidising, which rules out dilute nitric acid specifically.

Derivation AVolume of hydrogen gas from a given mass of metal

The goal is to turn a simple mass of a reactive metal into a predicted volume of hydrogen gas measured at STP, using nothing beyond the balanced equation's mole ratio, the metal's molar mass, and the standard molar volume of a gas at STP — a calculation examiners routinely attach to this reaction to test whether students can move from a word equation to a number.

Step 1. Write the balanced equation and read off the mole ratio between metal and hydrogen. For zinc reacting with dilute hydrochloric acid, $Zn + 2HCl \rightarrow ZnCl_2 + H_2\uparrow$ , one mole of zinc produces exactly one mole of hydrogen gas — the mole ratio of metal to $H_2$ is $1:1$ .

Step 2. Convert the given mass of metal to moles using its molar mass. Moles of metal $= \dfrac{\text{mass given (g)}}{\text{molar mass (g/mol)}}$ . This step is the same regardless of which metal or acid is involved; only the molar mass changes.

Step 3. Apply the mole ratio from the balanced equation to find moles of hydrogen produced. Because the ratio for zinc and hydrochloric acid is $1:1$ , moles of $H_2$ equal moles of zinc consumed; for a metal like aluminium, where the equation is $2Al + 3H_2SO_4 \rightarrow Al_2(SO_4)_3 + 3H_2\uparrow$ , the ratio of metal to $H_2$ is $2:3$ , so moles of $H_2 = \tfrac{3}{2} \times$ moles of $Al$ .

Step 4. Convert moles of hydrogen to volume using the molar volume of a gas at STP, $22{,}400\ \text{cm}^3$ per mole (equivalently $22.4\ \text{dm}^3\ \text{mol}^{-1}$ ):

V(H_2) = n(H_2) \times 22{,}400\ \text{cm}^3\text{mol}^{-1}

where: $n(H_2)$ is the number of moles of hydrogen gas found in Step 3, and $22{,}400\ \text{cm}^3$ is the volume occupied by one mole of any gas at standard temperature and pressure. The result is the volume of gas a student would expect to collect over water or in a gas syringe for that mass of metal.

Worked example

Question: 6.5 g of zinc granules are added to excess dilute hydrochloric acid. Calculate the volume of hydrogen gas liberated, measured at STP. (Molar mass of Zn = 65 g/mol; molar volume at STP = 22,400 cm³/mol.)

Step 1. Write the balanced equation: $Zn + 2HCl \rightarrow ZnCl_2 + H_2\uparrow$ . This is the same reaction confirmed earlier — zinc lies above hydrogen in the activity series, so effervescence is expected, and the mole ratio of $Zn$ to $H_2$ is $1:1$ .

Step 2. Convert the given mass of zinc to moles: moles of $Zn = \dfrac{6.5\ \text{g}}{65\ \text{g/mol}} = 0.1\ \text{mol}$ .

Step 3. Apply the $1:1$ mole ratio: moles of $H_2$ produced $= 0.1\ \text{mol}$ , the same as the moles of zinc consumed.

Step 4. Convert to volume at STP: $V(H_2) = 0.1\ \text{mol} \times 22{,}400\ \text{cm}^3\text{mol}^{-1} = 2240\ \text{cm}^3$ .

The reaction liberates 2240 cm³ (2.24 dm³) of hydrogen gas at STP.

Competitive stretch

If the same 6.5 g of metal were copper instead of zinc, no calculation is even needed — copper lies below hydrogen in the activity series and does not react with dilute hydrochloric acid at all, so the volume of hydrogen liberated is 0 cm³. Always check the activity-series position before reaching for the mole-ratio arithmetic; a stoichiometry calculation on a non-reacting metal is a wasted answer.

Common traps

Jumping straight into a mole-ratio calculation without first confirming the metal actually reacts — check the activity-series position first; metals below hydrogen give zero gas, no matter the mass.
Using the wrong mole ratio for metals with valency other than 2, such as aluminium (ratio $2:3$ with $H_2SO_4$ ) — always read the ratio off the balanced equation, never assume $1:1$ .
Applying the "metal + dilute acid gives hydrogen" rule to dilute nitric acid — it reacts with reactive metals, but the oxidising acid consumes the hydrogen as it forms, so no $H_2$ gas is collected.
Forgetting to balance the salt formula by the metal's valency, e.g. writing $AlSO_4$ instead of the correctly balanced $Al_2(SO_4)_3$ .

Ask the tutor: Try giving the AI tutor a different metal and mass and ask it to walk through predicting the reaction and calculating the volume of hydrogen produced.

Why does iron react with dilute sulphuric acid but not visibly with dilute nitric acid to give hydrogen?

Iron is reactive enough — it sits above hydrogen in the activity series — so it does react with both acids. With dilute sulphuric acid, the displaced hydrogen escapes freely as gas, giving the familiar fizzing. With dilute nitric acid, a reaction still occurs, but the acid's strong oxidising power intercepts the newly-formed hydrogen and converts it to water before it can bubble off, while the nitrogen in the acid is reduced to nitrogen oxides. So the absence of hydrogen gas with nitric acid is about oxidation, not about iron being unreactive.

If copper does not react with dilute acids, why do old copper coins sometimes look corroded?

That corrosion is not caused by dilute acids displacing hydrogen — copper genuinely cannot do that, since it lies below hydrogen in the activity series. The green or dull coating on old copper coins usually comes from slow reactions with oxygen, moisture, and carbon dioxide in the air, forming compounds like copper carbonate, or from exposure to other chemicals entirely. It is a different kind of chemistry from the metal-plus-dilute-acid reaction this topic covers, so do not use coin corrosion as evidence that copper reacts with dilute acids.

Does it matter whether I use dilute hydrochloric acid or dilute sulphuric acid for these reactions?

For the four prescribed metals — Mg, Al, Zn, Fe — both dilute HCl and dilute H2SO4 behave the same way: any metal above hydrogen liberates hydrogen gas with either acid, forming a chloride salt with HCl or a sulphate salt with H2SO4. The identity of the salt changes, but the underlying displacement chemistry, the activity-series rule, and the gas produced do not. The exception to watch for is not between these two acids but with dilute nitric acid, which behaves completely differently because of its oxidising nature.