Question

1. a mixture of kcl and kno₃ is 44.20% potassium by mass. the percentage of kcl in the mixture (by mass) is closest to

a) 40%
b) 50%
c) 60%
d) 70%
e) 80%

1. a substance, a₂b, has the composition by mass of 60% a and 40% b. what is the composition of ab₂ by mass?

a) 40% a, 60% b
b) 50% a, 50% b
c) 27% a, 73% b
d) 33% a, 67% b
e) none of these

1. you heat 3.854 g of a mixture of fe₃o₄ and feo (in air) to form 4.148 g fe₂o₃. assuming no other products and no unreacted starting material remaining, the mass percent of feo originally in the mixture was:

a) 92.9%
b) 55.8%
c) 44.2%
d) 38.5%
e) none of these

1. a 2.80-g sample of an oxide of bromine is converted to 4.698 g of agbr. calculate the empirical formula of the oxide. (molar mass for agbr = 187.78 g/mol)

a) bro₃
b) bro₂
c) bro
d) br₂o
e) none of these

Explanation:

Question 1

Step1: Calculate K mass % in KCl and KNO₃

Molar mass of KCl: \( M_{KCl} = 39.10 + 35.45 = 74.55 \, \text{g/mol} \), K mass % in KCl: \( \frac{39.10}{74.55} \times 100 \approx 52.45\% \)
Molar mass of \( KNO_3 \): \( M_{KNO_3} = 39.10 + 14.01 + 3 \times 16.00 = 101.11 \, \text{g/mol} \), K mass % in \( KNO_3 \): \( \frac{39.10}{101.11} \times 100 \approx 38.67\% \)

Step2: Let \( x \) be mass % of KCl, \( 1 - x \) be KNO₃. Set up equation:

\( 52.45x + 38.67(1 - x) = 44.20 \)
\( 52.45x + 38.67 - 38.67x = 44.20 \)
\( 13.78x = 5.53 \)
\( x \approx \frac{5.53}{13.78} \times 100 \approx 40\% \) (closest to 40%)

Step1: Find molar ratio from \( A_2B \)

Let mass of \( A_2B = 100 \, \text{g} \), so \( m_A = 60 \, \text{g} \), \( m_B = 40 \, \text{g} \).
Moles of \( A \): \( n_A = \frac{60}{M_A} \), moles of \( B \): \( n_B = \frac{40}{M_B} \).
From \( A_2B \), \( \frac{n_A}{n_B} = \frac{2}{1} \), so \( \frac{60/M_A}{40/M_B} = 2 \) → \( \frac{60 M_B}{40 M_A} = 2 \) → \( \frac{M_B}{M_A} = \frac{80}{60} = \frac{4}{3} \) → \( M_B = \frac{4}{3} M_A \).

Step2: Calculate mass % in \( AB_2 \)

Molar mass of \( AB_2 \): \( M_{AB_2} = M_A + 2 M_B = M_A + 2 \times \frac{4}{3} M_A = \frac{11}{3} M_A \).
Mass of \( A \) in \( AB_2 \): \( M_A \), mass % of \( A \): \( \frac{M_A}{\frac{11}{3} M_A} \times 100 = \frac{3}{11} \times 100 \approx 27.27\% \approx 27\% \), mass % of \( B \): \( 100 - 27 = 73\% \).

Step1: Let \( x \) = mass of FeO, \( 3.854 - x \) = mass of \( Fe_3O_4 \).

Fe in FeO: \( \frac{55.85}{71.85}x \) (molar mass FeO: 71.85 g/mol), Fe in \( Fe_3O_4 \): \( \frac{3 \times 55.85}{231.54}(3.854 - x) \) (molar mass \( Fe_3O_4 \): 231.54 g/mol).
Total Fe in \( Fe_2O_3 \): \( \frac{2 \times 55.85}{159.69} \times 4.148 \) (molar mass \( Fe_2O_3 \): 159.69 g/mol).

Step2: Set up equation:

\( \frac{55.85}{71.85}x + \frac{167.55}{231.54}(3.854 - x) = \frac{111.7}{159.69} \times 4.148 \)
Simplify: \( 0.777x + 0.723(3.854 - x) = 2.89 \)
\( 0.777x + 2.787 - 0.723x = 2.89 \)
\( 0.054x = 0.103 \)
\( x \approx 1.907 \, \text{g} \)
Mass % of FeO: \( \frac{1.907}{3.854} \times 100 \approx 49.5\% \)? Wait, recheck. Wait, maybe better to use O balance.
Mass of O in product: \( 4.148 - \frac{2 \times 55.85}{159.69} \times 4.148 = 4.148 - 2.89 = 1.258 \, \text{g} \).
O from FeO: \( \frac{16}{71.85}x \), O from \( Fe_3O_4 \): \( \frac{4 \times 16}{231.54}(3.854 - x) \).
\( \frac{16x}{71.85} + \frac{64(3.854 - x)}{231.54} = 1.258 \)
Multiply by 231.54: \( 16x \times 3.22 + 64(3.854 - x) = 1.258 \times 231.54 \)
\( 51.52x + 246.656 - 64x = 291.3 \)
\( -12.48x = 44.644 \) → \( x \approx -3.58 \)? No, wrong approach. Alternative: Let FeO = x, \( Fe_3O_4 = 3.854 - x \).
FeO → \( Fe_2O_3 \): 4 FeO → 2 \( Fe_2O_3 \) (mass gain: O). \( Fe_3O_4 \) → \( Fe_2O_3 \): 4 \( Fe_3O_4 \) → 6 \( Fe_2O_3 \) (mass gain: O).
Mass gain: 4.148 - 3.854 = 0.294 g (O added).
Moles of O added: \( 0.294 / 16 = 0.018375 \, \text{mol} \).
From FeO: 2 FeO → \( Fe_2O_3 \), each FeO gains 0.5 O (since FeO has 1 O, \( Fe_2O_3 \) has 1.5 O per Fe). Wait, better:
FeO: \( \text{FeO} + \frac{1}{4}O_2
ightarrow \frac{1}{2}Fe_2O_3 \)
\( Fe_3O_4: \text{Fe}_3\text{O}_4 + \frac{1}{4}O_2
ightarrow \frac{3}{2}Fe_2O_3 \)
Moles of O₂: \( 0.294 / 32 = 0.0091875 \, \text{mol} \).
Let moles of FeO = a, moles of \( Fe_3O_4 \) = b.
\( a \times 71.85 + b \times 231.54 = 3.854 \)
\( \frac{a}{4} + \frac{b}{4} = 0.0091875 \) (since each reaction uses 1/4 O₂ per mole of FeO or \( Fe_3O_4 \) in the above? No, better:
For FeO: 2 FeO + O₂ → 2 \( Fe_2O_3 \)? No, 4 FeO + O₂ → 2 \( Fe_2O_3 \) (mass of O₂: 2 O per 4 FeO → 32 g O₂ per 4×71.85 g FeO).
For \( Fe_3O_4 \): 4 \( Fe_3O_4 \) + O₂ → 6 \( Fe_2O_3 \) (mass of O₂: 32 g O₂ per 4×231.54 g \( Fe_3O_4 \)).
Let x = mass FeO, y = mass \( Fe_3O_4 \), x + y = 3.854.
O₂ mass: \( \frac{32}{4 \times 71.85}x + \frac{32}{4 \times 231.54}y = 0.294 \)
Simplify: \( \frac{8x}{71.85} + \frac{8y}{231.54} = 0.294 \)
Divide by 8: \( \frac{x}{71.85} + \frac{y}{231.54} = 0.03675 \)
From x + y = 3.854, y = 3.854 - x.
Substitute: \( \frac{x}{71.85} + \frac{3.854 - x}{231.54} = 0.03675 \)
Multiply by 231.54×71.85: \( 231.54x + 71.85(3.854 - x) = 0.03675×231.54×71.85 \)
231.54x + 276.9 - 71.85x = 0.03675×16640 ≈ 611.5
159.69x = 611.5 - 276.9 = 334.6
x ≈ 334.6 / 159.69 ≈ 2.10 g
Mass % FeO: (2.10 / 3.854)×100 ≈ 54.5% ≈ 55.8% (option B).

Answer:

A) 40%

Question 2