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  4. Some Basic Concepts of Chemistry Chapter-Wise Test 19

Some Basic Concepts Of Chemistry Chapter-Wise Test 19

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What volume of \( \ce{O2} \) at STP is required to burn 5.6 g of \( \ce{C2H4} \) completely to \( \ce{CO2} \) and \( \ce{H2O} \)? (Molar mass: \( \ce{C2H4} \) = 28 g/mol)

Reaction: \( \ce{C2H4 + 3O2 -> 2CO2 + 2H2O} \).

Moles of \( \ce{C2H4} \) = \( \frac{5.6}{28} \) = 0.2 mol.

1 mol \( \ce{C2H4} \) needs 3 mol \( \ce{O2} \); 0.2 mol needs 0.6 mol.

Volume = 0.6 × 22.4 = 13.44 L.

13.44 L
6.72 L
22.4 L
4.48 L
1

How many grams of \( \ce{Al2O3} \) are required to produce 10.8 g of \( \ce{Al} \) with excess \( \ce{CO} \)? (Molar masses: \( \ce{Al2O3} \) = 102 g/mol, Al = 27 g/mol)

Reaction: \( \ce{Al2O3 + 3CO -> 2Al + 3CO2} \).

Moles of Al = \( \frac{10.8}{27} \) = 0.4 mol.

2 mol Al from 1 mol \( \ce{Al2O3} \); 0.4 mol from 0.2 mol.

Mass = 0.2 × 102 = 20.4 g.

10.2 g
30.6 g
15.3 g
20.4 g
4

What is the mass of \( \ce{CaCl2} \) produced when 25 g of \( \ce{Ca(OH)2} \) reacts with excess \( \ce{HCl} \)? (Molar masses: \( \ce{Ca(OH)2} \) = 74 g/mol, \( \ce{CaCl2} \) = 111 g/mol)

Reaction: \( \ce{Ca(OH)2 + 2HCl -> CaCl2 + 2H2O} \).

Moles of \( \ce{Ca(OH)2} \) = \( \frac{25}{74} \) ≈ 0.3378 mol.

1 mol \( \ce{Ca(OH)2} \) produces 1 mol \( \ce{CaCl2} \); mass = 0.3378 × 111 ≈ 37.5 g.

18.75 g
37.5 g
55.5 g
74 g
2

What volume of \( \ce{CO2} \) at STP is produced when 9 g of \( \ce{C3H8} \) is burned completely? (Molar mass: \( \ce{C3H8} \) = 44 g/mol)

Reaction: \( \ce{C3H8 + 5O2 -> 3CO2 + 4H2O} \).

Moles of \( \ce{C3H8} \) = \( \frac{9}{44} \) ≈ 0.2045 mol.

1 mol \( \ce{C3H8} \) produces 3 mol \( \ce{CO2} \); 0.2045 mol produces 0.6135 mol.

Volume = 0.6135 × 22.4 ≈ 13.74 L.

6.87 L
20.61 L
9.16 L
13.74 L
4

A 2.5 L sample of a gas at STP has a mass of 5 g. If it contains only nitrogen and oxygen, what is its empirical formula? (Atomic masses: N = 14, O = 16)

Moles = \( \frac{2.5}{22.4} \) ≈ 0.1116 mol; molar mass = \( \frac{5}{0.1116} \) ≈ 44.8 g/mol.

Assume \( \ce{N_xO_y} \): \( 14x + 16y = 44.8 \).

Simplest ratio: \( \ce{N2O} \) (14 × 2 + 16 = 44); empirical formula = \( \ce{N2O} \).

\( \ce{N2O} \)
\( \ce{NO} \)
\( \ce{NO2} \)
\( \ce{N2O4} \)
1

How many grams of \( \ce{K2SO4} \) are produced when 11.2 g of \( \ce{KOH} \) reacts with excess \( \ce{H2SO4} \)? (Molar masses: \( \ce{KOH} \) = 56 g/mol, \( \ce{K2SO4} \) = 174 g/mol)

Reaction: \( \ce{2KOH + H2SO4 -> K2SO4 + 2H2O} \).

Moles of \( \ce{KOH} \) = \( \frac{11.2}{56} \) = 0.2 mol.

2 mol \( \ce{KOH} \) produce 1 mol \( \ce{K2SO4} \); 0.2 mol produce 0.1 mol.

Mass = 0.1 × 174 = 17.4 g.

8.7 g
17.4 g
34.8 g
13.05 g
1

A solution contains 30 ppm of \( \ce{Na2SO4} \) by mass in water. What is its molarity if the density is 1 g/mL? (Molar mass: \( \ce{Na2SO4} \) = 142 g/mol)

30 ppm = 30 g \( \ce{Na2SO4} \) in 10⁶ g solution.

Volume = \( \frac{10⁶}{1} \) = 10⁶ mL = 1000 L.

Moles = \( \frac{30}{142} \) ≈ 0.2113 mol; molarity = \( \frac{0.2113}{1000} \) ≈ 2.113 × 10⁻⁴ M.

2.0 × 10⁻⁴ M
2.5 × 10⁻⁴ M
2.113 × 10⁻⁴ M
1.5 × 10⁻⁴ M
3

What is the mass of \( \ce{FeCl2} \) produced when 11.2 g of \( \ce{Fe} \) reacts with excess \( \ce{HCl} \)? (Molar masses: Fe = 56 g/mol, \( \ce{FeCl2} \) = 126.5 g/mol)

Reaction: \( \ce{Fe + 2HCl -> FeCl2 + H2} \).

Moles of Fe = \( \frac{11.2}{56} \) = 0.2 mol.

1 mol Fe produces 1 mol \( \ce{FeCl2} \); mass = 0.2 × 126.5 = 25.3 g.

12.65 g
25.3 g
50.6 g
18.9 g
2

A compound contains 27.27% carbon and 72.73% oxygen by mass. What is its empirical formula? (Atomic masses: C = 12, O = 16)

For 100 g: C = 27.27 g, O = 72.73 g.

Moles: C = \( \frac{27.27}{12} \) ≈ 2.27, O = \( \frac{72.73}{16} \) ≈ 4.55.

Ratio = 1 : 2; empirical formula = \( \ce{CO2} \).

\( \ce{CO} \)
\( \ce{CO2} \)
\( \ce{C2O4} \)
\( \ce{C2O} \)
2

What is the mass percentage of \( \ce{KNO3} \) in a solution made by dissolving 10.1 g of \( \ce{KNO3} \) in 40 g of water? (Molar mass: \( \ce{KNO3} \) = 101 g/mol)

Total mass = 10.1 + 40 = 50.1 g.

Mass % = \( \frac{10.1}{50.1} \) × 100 ≈ 20.16%.

25%
15%
20.16%
10%
3

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