20 cm³ of an aq 1.0 mol/dm³ solution of the hydroxide of metal X requires 40 cm³ of aq 0.25 mol/dm³ of sulfuric acid for complete neutralisation.

 

What is the formula for the sulfate of X?

 

A)  X₂SO₄

 

B)  XSO₄

 

C)  X₂(SO₄)₃

 

D)  X(SO₄)₂

 

 

Answer: A

 

Approach: calculate the no. of moles of X hydroxide and sulfuric acid and then compare the mole ratio.

 

By comparing mole ratio: 2 mol of X hydroxide reacted with 1 mol of sulfuric acid → X hydroxide must be XOH (monobasic).

 

2XOH + H₂SO₄ → X₂SO₄ + 2H₂O

 

Essentially, neutralisation is represented by H⁺ + OH^– → H₂O
i.e. 1 mol H⁺ reacts with 1 mol OH^–, hence:

 

one mol of monobasic acid (one H⁺) will react with one mol monobasic base (one OH^–)
one mol of dibasic acid (two H⁺)will react with two mol monobasic base (two OH^–)
two mol of monobasic acid (two H⁺) will react with one mol dibasic base (two OH^–)
...

 

All ammonium salts on heating with sodium hydroxide produce ammonia gas.

 

For which ammonium salt can the greatest mass of ammonia be obtained?

 

A)  0.5 mol (NH₄)₃PO₄

 

B)  0.5 mol (NH₄)₂SO₄

 

C)  1.0 mol NH₄Cl

 

D)  1.0 mol NH₄NO₃

 

 

Answer: A

 

Ionic eqn: NH₄⁺ + OH⁻ → NH₃ + H₂O
→ 1 mol NH₄⁺ will produce 1 mol NH₃

 

- the larger the no. of moles of NH₄⁺ present, the larger the no. of moles of NH₃ produced, the greater the mass of ammonia obtained.

 

No. of moles of NH₄⁺ present in:
A = 0.5 x 3 = 1.5 mol
B = 0.5 x 2 = 1.0 mol
C = 1 x 1 = 1.0 mol
D = 1 x 1 = 1.0 mol

 

Zinc nitrate crystals decompose on heating according to the following equation:

 

 2Zn(NO₃)₂.6H₂O (s) → 2ZnO (s) + 4NO₂ (g) + O₂ (g) + 12H₂O (g)

 

If 14.85 g of zinc nitrate crystals are heated strongly and then allowed to cool, what is the total volume of gas obtained at room temperature and pressure?

 

A)  2.4 dm³

 

B)  3.0 dm³

 

C)  5.1 dm³

 

D)  10.2 dm³

 

 

Answer: B

 

- At room temp and pressure, water is no longer a gas
→ 2 moles of Zn(NO₃)₂.6H₂O will produce only 5 moles of gases.

 

M_r of Zn(NO₃)₂.6H₂O = 297
No. of moles of Zn(NO₃)₂.6H₂O = 14.85/297 = 0.05 mol
No. of moles of gases = 5/2 x 0.05 = 0.125 mol
Volume of gases = 0.125 x 24 = 3.0 dm³

 

In the atmosphere, nitrogen can undergo a series of reactions to form nitric acid:

 

 N₂ + O₂ → 2NO
 2NO + O₂ → 2NO₂
 4NO₂ + 2H₂O → 4HNO₃

 

How many moles of nitric acid are formed from 0.25 moles of nitrogen?

 

A)  0.25

 

B)  0.50

 

C)  1.00

 

D)  4.00

 

 

Answer: B

 

Simplest approach:

 

Use reacting ratio for each equation one at a time i.e.

 

N₂:NO = 1:2
→ 0.25 mol N₂ gives 0.50 mol NO

 

NO:NO₂ = 2:2
→ 0.50 mol NO gives 0.50 mol NO₂

 

NO₂:HNO₃ = 4:4
→ 0.50 mol NO₂ gives 0.50 mol HNO₃

 

A sample of urine containing 0.120 g of urea, NH₂CONH₂, was treated with an excess of nitrous acid. The acid reacted according to the following equation:

 

 H₂ + 2HNO₂ → CO₂ + 2N₂ + 3H₂O

 

The gas produced was passed through aqueous sodium hydroxide and the final volume measured.

 

What was this volume at room temperature and pressure?

 

A)  9.6 cm³

 

B)  14.4 cm³

 

C)  48.0 cm³

 

D)  96.0 cm³

 

 

Answer: D

 

- At room temperature and pressure, H₂O will be a liquid.
- When the gases pass through aq NaOH (strong alkali), the acidic gases will be removed i.e. CO₂ will be removed.
- The only gas remaining is N₂.

 

M_r of urea = 60
No. of moles of urea = 0.120/60 = 2 x 10^–3 mol

 

Based on reacting ratio,
No. of moles of N₂ = 2 x 2 x 10^–3 = 4 x 10^–3 mol

 

Volume of N₂ = 4 x 10^–3 x 24000 cm³ = 96 cm³

 

When charcoal reacts in the presence of oxygen, carbon monoxide and carbon dioxide are produced according to the following chemical reactions.

 

 2C(s) + O₂(g) → 2CO(g)

 

 C(s) + O₂(g) → CO₂(g)

 

What would be the total mass of gas produced when 400 g of charcoal is reacted, assuming equal amounts are consumed in each reaction?

 

A)  0.93 kg

 

B)  1.2 kg

 

C)  1.5 kg

 

D)  2.5 kg

 

 

Answer: B

 

- Equal amounts are consumed i.e. 200 g of C undergoes each reaction

 

No. of moles of C reacted in each reaction = 200/12 = 16.67 mol

 

No. of moles of CO formed = 16.67 mol
Mass of CO formed = 16.67 x (12+16) = 466.67 g

 

No. of moles of CO₂ formed = 16.67 mol
Mass of CO₂ formed = 16.67 x (12+16x2) = 733.33 g

 

Total mass of gas formed = 1200 g = 1.2 kg

 

A compound Y is the only substance formed when two volumes of dry ammonia gas react with one volume of dry carbon dioxide (both volumes measured at r.t.p.).

 

What is the most likely formula of Y?

 

A)  (NH₄)₂CO₃

 

B)  NH₂COONH₄

 

C)  (NH₂)₂CO

 

D)  NH₄COONH₄

 

 

Answer: B

 

- In a reaction involving gases, volume ratio = molar ratio (Avogadro's relation)

 

- Since 2NH₃ + CO₂ → Y, the molecular formula of Y must contain 2N, 6H, 1C and 2O.