Answers · Concentration, gas volumes and yield

When does the acid run out? — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. YQ10 · Practice

Model answer. The limiting reactant is used up completely, so the reaction stops when it runs out and it decides how much product forms. The other reactant is in excess: some of it is left over.

A full answer includes

The reactant that is used up completely; it decides how much product forms.

Also accept

The reactant used up completely

Watch for

“The limiting reactant is the one with fewer moles.” — Divide each amount by its number in the equation first. In Mg + 2HCl, the acid can run out even when it has more moles, because each Mg needs two HCl.
“The limiting reactant is the one left over.” — The one left over is in excess. The limiting reactant is all used up.

2. YQ11 · Practice

Model answer. Magnesium is left, so it cannot have run out. The fizzing stops because the acid is used up: the acid is the limiting reactant and the magnesium is in excess. The 100 cm³ syringe was not full.

A full answer includes

The acid has run out, so it is the limiting reactant; the magnesium left over is in excess.

Also accept

The acid ran out (limiting); magnesium in excess

Watch for

“The syringe was full.” — The syringe holds 100 cm³ and stopped at 60 cm³: the gas stopped coming.
“The magnesium ran out.” — There are pieces of magnesium left in the flask, so it has not run out.

3. YQ12 · Practice

Model answer. n(Fe) = 2.80 ÷ 56 = 0.0500 mol and n(S) = 2.00 ÷ 32 = 0.0625 mol. The ratio is 1 : 1, so iron runs out first: it is limiting. 0.0500 mol of FeS forms; M(FeS) = 88 g/mol, so m = 0.0500 × 88 = 4.40 g, and some sulfur is left over.

Checked automatically

(a) Amount of iron: 0.0500 mol (3 significant figures)
(b) Amount of sulfur: 0.0625 mol (3 significant figures)
(c) Mass of iron(II) sulfide: 4.40 g (3 significant figures)
- Common wrong answer 5.50 : You used the sulfur. 0.0625 mol of sulfur would need 0.0625 mol of iron, and there is only 0.0500 mol: iron is limiting.
- Common wrong answer 4.80 : You added the masses. Some sulfur is left over, so the product is lighter.

A full answer includes

(a) 2.80 ÷ 56 = 0.0500 mol.
(b) 2.00 ÷ 32 = 0.0625 mol.
(c) Iron is limiting (0.0500 < 0.0625): 0.0500 mol of FeS × 88 = 4.40 g.

Also accept

0.0500 mol; 0.0625 mol; 4.40 g, iron limiting

Watch for

“The product has the mass of both reactants added together.” — Only if both are used up. The reactant in excess is partly left over, so use the limiting reactant.

4. YQ13 · Practice

Model answer. n(Mg) = 0.0200 mol and n(HCl) = 1.00 × 0.0250 = 0.0250 mol. Dividing by the numbers in the equation gives 0.0200 for Mg and 0.0125 for HCl, so the acid is limiting even though it has more moles. Two HCl make one H₂, so 0.0125 mol of hydrogen forms: 0.0125 × 24 000 = 300 cm³. The same 0.0125 mol of magnesium chloride is dissolved in 25.0 cm³ = 0.0250 dm³, so its concentration is 0.0125 ÷ 0.0250 = 0.500 mol/dm³.

Checked automatically

(b) Volume of hydrogen: 300 cm³ (3 significant figures)
- Common wrong answer 480 : You used the magnesium. 0.0200 mol of Mg needs 0.0400 mol of HCl, and there is only 0.0250 mol: the acid is limiting.
- Common wrong answer 600 : Two HCl make one H₂, so 0.0250 mol of HCl makes 0.0125 mol of hydrogen.
(c) Concentration of magnesium chloride: 0.500 mol/dm³ (3 significant figures)
- Common wrong answer 0.800 : You used the magnesium. The acid is limiting, so only 0.0125 mol of MgCl₂ forms.
- Common wrong answer 0.000500 : Change 25.0 cm³ to dm³ by dividing by 1000: 0.0250 dm³. Then c = n ÷ V.
- Common wrong answer 1.00 : Two HCl make one MgCl₂, so 0.0250 mol of HCl makes 0.0125 mol of MgCl₂.

A full answer includes

n(Mg) = 0.480 ÷ 24 = 0.0200 mol; n(HCl) = 1.00 × 0.0250 = 0.0250 mol.
(a) Mg 0.0200 ÷ 1 = 0.0200; HCl 0.0250 ÷ 2 = 0.0125: the acid is limiting.
(b) 0.0250 mol of HCl makes 0.0125 mol of H₂: 0.0125 × 24 000 = 300 cm³.
(c) It also makes 0.0125 mol of MgCl₂ in 0.0250 dm³: c = 0.0125 ÷ 0.0250 = 0.500 mol/dm³.

Also accept

(a) the acid (b) 300 cm³ (c) 0.500 mol/dm³

Watch for

“The limiting reactant is the one with fewer moles.” — Divide each amount by its number in the equation first. In Mg + 2HCl, the acid can run out even when it has more moles, because each Mg needs two HCl.

5. YQ14 · Practice

Model answer. n(Mg) = 0.300 ÷ 24 = 0.0125 mol and n(HCl) = 0.0100 mol, enough for only 0.00500 mol of magnesium: the acid is limiting. 0.00500 × 24 = 0.120 g of magnesium reacts, so 0.300 − 0.120 = 0.180 g is left.

Checked automatically

Mass of magnesium left: 0.180 g (3 significant figures)
- Common wrong answer 0.120 : That is the magnesium that reacts. The question asks what is left: 0.300 − 0.120 g.
- Common wrong answer 0.0600 : Each Mg needs two HCl, so 0.0100 mol of HCl reacts with only 0.00500 mol of magnesium, 0.120 g.
- Common wrong answer 0.240 : Each Mg needs two HCl, so 0.0100 mol of HCl reacts with only 0.00500 mol of magnesium, 0.120 g. Then subtract it from 0.300 g.
- Common wrong answer 0 : The acid holds 0.0100 mol of HCl, enough for only 0.00500 mol of magnesium.

A full answer includes

n(HCl) = 0.500 × 0.0200 = 0.0100 mol; it reacts with 0.0100 ÷ 2 = 0.00500 mol of Mg.
That is 0.00500 × 24 = 0.120 g of magnesium.
Left: 0.300 − 0.120 = 0.180 g.

Also accept

0.180 g

Watch for

“The limiting reactant is the one with fewer moles.” — Divide each amount by its number in the equation first. In Mg + 2HCl, the acid can run out even when it has more moles, because each Mg needs two HCl.

6. YQ15 · Practice

Model answer. n(BaCl₂) = 0.00200 mol and n(Na₂SO₄) = 0.00100 mol. In a 1 : 1 ratio the smaller amount, sodium sulfate, is limiting, so 0.00100 mol of barium sulfate forms. M(BaSO₄) = 137 + 32 + 4 × 16 = 233 g/mol, so m = 0.00100 × 233 = 0.233 g.

Checked automatically

Mass of barium sulfate: 0.233 g (3 significant figures)
- Common wrong answer 0.466 : You used the barium chloride. Sodium sulfate has fewer moles in a 1 : 1 ratio, so it is limiting.

A full answer includes

n(BaCl₂) = 0.100 × 0.0200 = 0.00200 mol; n(Na₂SO₄) = 0.100 × 0.0100 = 0.00100 mol.
The ratio is 1 : 1, so sodium sulfate is limiting: 0.00100 mol of BaSO₄.
M(BaSO₄) = 233 g/mol; m = 0.00100 × 233 = 0.233 g.

Also accept

0.233 g

Watch for

“The product has the mass of both reactants added together.” — Only if both are used up. The reactant in excess is partly left over, so use the limiting reactant.

7. YQ16 · Practice

Model answer. The copper(II) oxide is in excess, because some is left over. Adding it in excess makes sure that all the sulfuric acid is used up, so the solution holds only the salt and water, and the extra solid can be filtered off.

A full answer includes

(a) Copper(II) oxide: the black solid left over.
(b) So that all the acid reacts: none is left in the solution with the salt.
The solid left over is easy to remove by filtering.

Also accept

(a) copper(II) oxide (b) all the acid reacts; the extra solid is filtered off

Watch for

“The excess makes more salt.” — The acid is limiting, so it decides how much salt forms. The excess makes sure that all of the acid reacts.

8. YT2 · Apply

Model answer. For gases the volumes come in the mole ratio. Divide each volume by its number in the equation: N₂ 100 ÷ 1 = 100, H₂ 240 ÷ 3 = 80. Hydrogen gives the smaller number, so it is limiting. It makes 240 × 2/3 = 160 cm³ of ammonia and uses 80 cm³ of nitrogen, so 20 cm³ of nitrogen is left over.

Checked automatically

(a) Volume of ammonia: 160 cm³ (exact value)
- Common wrong answer 200 : You used the nitrogen. 100 cm³ of N₂ would need 300 cm³ of hydrogen, and there is only 240 cm³: hydrogen is limiting.
- Common wrong answer 480 : H₂ : NH₃ is 3 : 2, so the ammonia is two thirds of the hydrogen.
(b) Volume of gas left over: 20 cm³ (exact value)

A full answer includes

100 cm³ of N₂ would need 300 cm³ of H₂; there is only 240 cm³, so hydrogen is limiting.
(a) NH₃ = 240 × 2/3 = 160 cm³.
(b) N₂ used = 240 ÷ 3 = 80 cm³; 100 − 80 = 20 cm³ of nitrogen left.

Also accept

160 cm³ of ammonia; 20 cm³ of nitrogen left

Watch for

“The limiting reactant is the one with fewer moles.” — Divide each amount by its number in the equation first. In Mg + 2HCl, the acid can run out even when it has more moles, because each Mg needs two HCl.