Concentration, gas volumes and yield — return questions key

1. YR1 · Return

Model answer. One mole of any gas takes up 24 dm³ at r.t.p. Both samples hold one mole, the same number of particles. In a gas the particles are far apart, so the volume is almost all space between them: the number of particles decides it, not their mass or size.

A full answer includes

• (a) 24 dm³ (24 000 cm³).
• (b) Each holds the same number of particles: one mole, 6.02 × 10²³.
• Gas particles are far apart, so the volume is mostly space between them and the size or mass of the particles hardly matters.

Also accept

• (a) 24 dm³ (b) same number of particles; particles far apart

Watch for

• “Heavier particles take up more room.” — In a gas the particles are far apart, so their own size and mass hardly change the volume.

2. YR2 · Return

Model answer. M(Fe₂O₃) = 2 × 56 + 3 × 16 = 160 g/mol, so n = 16.0 ÷ 160 = 0.100 mol. One Fe₂O₃ gives two Fe, so 0.200 mol of iron: 0.200 × 56 = 11.2 g.

Checked automatically

• Mass of iron: 11.2 g (3 significant figures)
  • Common wrong answer 5.60 : One Fe₂O₃ gives two Fe: double the moles, 0.200 mol.
  • Common wrong answer 16.0 : The mass does not stay the same from reactant to product. Go through moles.

A full answer includes

• M(Fe₂O₃) = 160 g/mol; n = 16.0 ÷ 160 = 0.100 mol.
• Fe₂O₃ : Fe = 1 : 2, so 0.200 mol of Fe.
• m = 0.200 × 56 = 11.2 g.

Also accept

• 11.2 g

Watch for

• “The product has the same mass as the reactant.” — Only the total mass is conserved. Go through moles: change to moles, use the equation, change back.

3. YR3 · Return

Model answer. In the first runs the magnesium is limiting and all of it reacts, so the volume rises in proportion to the mass. The acid holds 0.500 × 0.0120 = 0.00600 mol of HCl, enough for 0.00300 mol of hydrogen, 72 cm³. Once the magnesium is more than enough, the acid is limiting: it runs out at 72 cm³ and the extra magnesium is left in excess.

A full answer includes

• (a) At first the magnesium is limiting and the acid is in excess: all the magnesium reacts, so more magnesium gives more hydrogen, in proportion.
• (b) Then the acid is limiting: it is all used up, so no more hydrogen forms, and the extra magnesium is left over (in excess).

Also accept

• (a) magnesium limiting, all reacts (b) acid limiting, used up; magnesium in excess

Watch for

• “The syringe or the reaction slowed down.” — The volume stops because a reactant, the acid, has run out. Magnesium is left over.

4. YR4 · Return

Model answer. n(CaCO₃) = 0.100 mol, and one CaCO₃ gives one CaO, so the theoretical yield is 0.100 × 56 = 5.60 g. Percentage yield = 5.04 ÷ 5.60 × 100 = 90.0 %. Calcium is 40 of the 56 in one CaO: 40 ÷ 56 × 100 = 71.4 % by mass.

Checked automatically

• (a) Theoretical yield: 5.60 g (3 significant figures)
• (b) Percentage yield: 90.0 % (3 significant figures)
  • Common wrong answer 50.4 : You divided by the mass of calcium carbonate. Divide by the theoretical yield of calcium oxide, 5.60 g.
• (c) Calcium in calcium oxide: 71.4 % (3 significant figures)
  • Common wrong answer 40.0 : You used calcium carbonate, M = 100. The question asks about calcium oxide, M = 56.
  • Common wrong answer 28.6 : That is the oxygen. Put the calcium on top: 40 ÷ 56 × 100.

A full answer includes

• (a) n(CaCO₃) = 10.0 ÷ 100 = 0.100 mol, so 0.100 mol of CaO: 0.100 × 56 = 5.60 g.
• (b) 5.04 ÷ 5.60 × 100 = 90.0 %.
• (c) Ca is 40 of the 56 in CaO: 40 ÷ 56 × 100 = 71.4 %.

Also accept

• 5.60 g; 90.0 %; 71.4 %

Watch for

• “Percentage yield = theoretical yield ÷ actual yield × 100.” — Put the mass you actually collected on top: actual ÷ theoretical × 100. A dry, pure product never gives more than 100 %.

5. YR5 · Return

Model answer. In 100 g there are 40.0 g of sulfur and 60.0 g of oxygen. S: 40.0 ÷ 32 = 1.25 mol; O: 60.0 ÷ 16 = 3.75 mol. Dividing by 1.25 gives 1 : 3: SO₃.

A full answer includes

• S = 40.0 ÷ 32 = 1.25 mol; O = 60.0 ÷ 16 = 3.75 mol.
• S : O = 1 : 3, so SO₃.

Also accept

• SO₃

Watch for

• “Use 40 : 60 = 2 : 3 as the formula ratio.” — That is a ratio of masses. Divide each by its Ar to count atoms.