Honors Chemistry

Answers to Chapter 3 Study Questions

1. a) 3(12.0) + 8(1.01) + 3(16.0) = 92.1 g/mole b) 92.1 g c) 6.02 x 10²³ molecules

d) = 20.0 g

2. a) 4 atoms (one N + 3 H)

b) = 2.41 x 10²⁴ atoms

c) = 4.82 x 10²³ atoms

3. Molar mass of NaNO₂ = 23.0 + 14.0 + 2(16.0) = 69.0 g/mole

% Na = 23.0/69.0 = 33.3% Na; % N = 14.0/69.0 = 20.3% N; %O = 2(16.0)/69.0 = 46.4% O

33.3% Na, 20.3% N and 46.4% O.

4. a) In 100 g of this compound, there are 40.7 g C, 5.1 g H, and 54.2 g O

= 3.39 moles C 3.39/3.39 = 1 x 2 = 2

= 5.1 moles H 5.1/3.39 = 1.5 x 2 = 3

= 3.39 moles O 3.39/3.39 = 1 x 2 = 2

empirical formula = C₂H₃O₂

b) Molar mass of C₂H₃O₂ = 2(12.0) + 3(1.0) + 2(16.0) = 59.0 g/mole

118/59.0 = 2 ® molecular formula = C₄H₆O₄

5. a) In 25.0 g of compound, there are 7.20 g Mg, 3.55 g C and 25.0-(7.20+3.55) = 14.25 g O.

= 0.296 moles Mg 0.296/0.296 = 1

= 0.296 moles C 0.296/0.296 = 1

= 0.891 moles O 0.891/0.296 = 3

formula = MgCO₃

b) % Mg = 7.20/25.0 = 28.8% Mg; %C = 3.55/25.0 = 14.2 % C; % O = 14.25/25.0 = 57.0% O. (You should get the same result using molar mass and atomic masses.)

c) = 4.00 g Mg

5. d) = 24.5 g compound

6. mass of Zn = 33.64 g - 32.00 g = 1.64 g Zn

mass of O = 34.04 g - 33.64 g = 0.40 g O.

# moles Zn: = 0.0251 moles Zn 0.251/0.25 = 1

# moles O: = 0.025 moles O 0.25/0.25 = 1

formula = ZnO

7. a) B₂H₆(l) + 3 O₂(g) ® B₂O₃(s) + 3 H₂O(l)

b) 4 PH₃(g) + 8 O₂(g) ® 6 H₂O(l) + P₄O₁₀(s)

8. a) 6 Li(s) + N₂(g) ® 2 Li₃N(s)

b) Co(NO₃)₃(aq) + 3 NaOH(aq) ® 3 NaNO₃(aq) + Co(OH)₃ (s)

c) Zn(s) + 2 HCl(aq) ® ZnCl₂(aq) + H₂(g)

d) (a) is a combination (synthesis) reaction. (b) is a double replacement reaction.

9. a) = 8.20 moles H₂S

b) = 48.0 g O₂

c) = 3.60 g H₂O

d) = 15.0 g SO₂

= 80.0%

e) = 5.00 g SO₂

= 4.01 g SO₂

O₂ is limiting; 4.01 g SO₂ is produced

10. a) H₂(g) + Cl₂(g) ® 2 HCl(g)

b) H₂ is limiting (H₂ and Cl₂ react in a ratio 1:1 ratio and there is less H₂.)

c) = 15.0 mol HCl

d) 9.00 - 7.50 = 1.50 mol Cl₂ left