Let’s break down the questions and perform calculations step by step for each reaction.
Question 41: \( 2Li + F_2 \rightarrow 2LiF \)
a. Show that 31 g of Lithium produces approximately 115 g of lithium fluoride.
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Molar mass of Li = 6.94 g/mol
- \( \text{Moles of Li} = \frac{31 \text{ g}}{6.94 \text{ g/mol}} \approx 4.47 \text{ moles of Li} \)
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According to the equation, 2 moles of Li produce 2 moles of LiF.
- Therefore, 4.47 moles of Li will produce 4.47 moles of LiF as well.
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Molar mass of LiF (Li = 6.94 + F = 19.00) = 25.94 g/mol
- \( \text{Mass of LiF} = 4.47 \text{ moles} \times 25.94 \text{ g/mol} \approx 115.06 \text{ g} \)
Conclusion: 31 g of Lithium produces approximately 115 g of lithium fluoride.
b. Show that 90 g of fluorine produces approximately 123 g of lithium fluoride.
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Molar mass of F2 = 38.00 g/mol (F = 19.00 g/mol, so F2 = 19.00*2)
- \( \text{Moles of F2} = \frac{90 \text{ g}}{38.00 \text{ g/mol}} \approx 2.37 \text{ moles of F2} \)
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From the equation, 1 mole of F2 produces 2 moles of LiF.
- Moles of LiF produced = 2 × 2.37 = 4.74 moles of LiF.
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\( \text{Mass of LiF} = 4.74 \text{ moles} \times 25.94 \text{ g/mol} \approx 122.99 \text{ g} \)
Conclusion: 90 g of fluorine produces approximately 123 g of lithium fluoride.
c. What mass of lithium was used if 87 g of lithium fluoride is produced?
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Calculate moles of LiF produced:
- \( \text{Moles of LiF} = \frac{87 \text{ g}}{25.94 \text{ g/mol}} \approx 3.35 \text{ moles of LiF} \)
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From the reaction \( 2LiF \) is produced from \( 2Li \), so moles of Li = 3.35 moles of LiF.
- Thus, \( \text{Mass of Li} = 3.35 \text{ moles} \times 6.94 \text{ g/mol} \approx 23.24 \text{ g} \)
Question 42: \( 2Ca + O_2 \rightarrow 2CaO \)
a. Show that approximately 107 g of calcium is needed to produce 150 g of calcium oxide.
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Molar mass of Ca = 40.08 g/mol
- Moles of CaO = \( \frac{150 \text{ g}}{56.08 \text{ g/mol}} \approx 2.67 \text{ moles} \)
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According to the reaction \( 2Ca \) produces \( 2CaO \), moles of Ca needed = 2.67 moles.
- So, moles of Ca = 2.67 moles.
- Mass of Ca = \( 2.67 \times 40.08 \approx 107.06 \text{ g} \)
Conclusion: Approximately 107 g of calcium is needed to produce 150 g of calcium oxide.
b. Show that the mass of calcium oxide produced from 1 kg of calcium is approximately 1400 g. (change everything into grams!)
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1 kg of Ca = 1000 g.
- Moles of Ca = \( \frac{1000 \text{ g}}{40.08 \text{ g/mol}} \approx 24.92 \text{ moles} \)
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From \( 2Ca + O_2 \) produces \( 2CaO \), 24.92 moles of Ca will produce 24.92 moles of CaO.
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Mass of CaO = \( 24.92 \text{ moles} \times 56.08 \text{ g/mol} \approx 1399.51 \text{ g} \)
Conclusion: Approximately 1400 g of calcium oxide formed from 1 kg of calcium.
c. Calculate the mass of calcium oxide formed from 0.82 g of oxygen.
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Molar mass of O2 = 32.00 g/mol
- Moles of O2 = \( \frac{0.82}{32.00} \approx 0.025625 \text{ moles of O2} \)
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From the reaction, 1 mole of O2 is required for 2 moles of CaO.
- So, moles of CaO produced = \( 0.025625 \times 2 \approx 0.05125 \text{ moles} \)
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Mass of CaO = \( 0.05125 \times 56.08 \approx 2.87 \text{ g} \)
Question 43: \( P_4 + 5O_2 \rightarrow 2P_2O_5 \)
a. Show that the amount of product formed from 43g of phosphorus is just less than 100g.
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Molar mass of P4 = 123.88 g/mol
- Moles of P4 = \( \frac{43 \text{ g}}{123.88 \text{ g/mol}} \approx 0.3472 \text{ moles of P4} \)
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According to the reaction, \( 1 \text{ mole of P4 produces 2 moles of P2O5} \)
- Moles of P2O5 = \( 0.3472 \times 2 = 0.6944 \text{ moles} \)
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Molar mass of P2O5 = 142.00 g/mol
- Mass of P2O5 = \( 0.6944 \times 142.00 \approx 98.87 \text{ g} \)
Conclusion: The amount of product formed from 43g of phosphorus is just less than 100g.
b. 21g of oxygen is used. How much product is formed?
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Molar mass of O2 = 32.00 g/mol
- Moles of O2 = \( \frac{21 \text{ g}}{32.00 \text{ g/mol}} = 0.65625 \text{ moles} \)
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From the reaction, \( 5 \text{ moles of O2 produce 2 moles of P2O5} \)
- Moles of P2O5 produced = \( \frac{0.65625}{5} \times 2 = 0.263 \text{ moles} \)
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Mass of P2O5 = \( 0.263 \times 142.00 \approx 37.46 \text{ g} \)
Question 44: \( 4Al + 3Cl2 \rightarrow 2Al2Cl3 \)
a. 1.3kg of aluminium is used. How much aluminium chloride is produced?
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Convert 1.3kg to grams: 1300g.
- Moles of Al = \( \frac{1300 \text{ g}}{26.98 \text{ g/mol}} \approx 48.24 \text{ moles} \)
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From the reaction, 4 moles of Al produce 2 moles of Al2Cl3
- Moles of Al2Cl3 produced = \( \frac{48.24}{4} \times 2 \approx 24.12 \text{ moles} \)
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Molar mass of Al2Cl3 = 133.33 g/mol
- Mass of Al2Cl3 = \( 24.12 \times 133.33 \approx 3216.01 \text{ g} \)
Conclusion: Approximately 3216.01 g of Aluminium chloride is produced.
b. How much chlorine is required to react with 77g of aluminium?
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Moles of Al = \( \frac{77}{26.98} \approx 2.86 \text{ moles} \)
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According to the reaction, 4 moles of Al are needed for 3 moles of Cl2.
- Moles of Cl2 needed = \( \frac{2.86}{4} \times 3 \approx 2.14 \text{ moles} \)
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Mass of Cl2 = \( 2.14 \times 70.91 \approx 151.33 \text{ g} \)
Question 45: \( 2K + 2H2O \rightarrow 2KOH + H2 \)
a. How much hydrogen is produced from a reaction using 132g of water?
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Molar mass of H2O = 18.02 g/mol
- Moles of H2O = \( \frac{132}{18.02} \approx 7.33 \text{ moles} \)
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According to the reaction, 2 moles of water produce 1 mole of H2.
- Moles of H2 = \( \frac{7.33}{2} \approx 3.67 \text{ moles} \)
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Mass of H2 = \( 3.67 \times 2.02 \approx 7.40 \text{ g} \)
b. How much potassium is required to produce 0.55g of potassium hydroxide?
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Molar mass of KOH = 56.11 g/mol
- Moles of KOH = \( \frac{0.55}{56.11} \approx 0.00980 \text{ moles} \)
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According to the reaction, 2 moles of KOH are produced from 2 moles of K.
- Moles of K needed = \( 0.00980 \text{ moles} \)
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Mass of K = \( 0.00980 \times 39.10 \approx 0.383 \text{ g} \)
c. Suggest why the mass of this reaction appears to decrease.
The apparent decrease in mass is due to the production of hydrogen gas (\( H_2 \)), which escapes into the atmosphere during the reaction.
Question 46: Challenge: Combustion of \( C_4H_{10} \)
For the complete combustion: \[ C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O \]
For the incomplete combustion: \[ C_4H_{10} + 6O_2 \rightarrow 4CO + 5H_2O \]
- From 15g of Butane (\(C_4H_{10}\)):
- Molar mass of \( C_4H_{10} \) = 58.12 g/mol
- Moles = \( \frac{15}{58.12} \approx 0.258 \text{ moles} \)
Complete Combustion:
From the complete reaction, 1 mole of \( C_4H_{10} \) reacts with 13 moles of \( O_2 \).
- \( O_2 \) needed = \( 0.258 \times 13 \approx 3.36 \text{ moles of } O_2 \)
Incomplete Combustion:
From the incomplete reaction, 1 mole of \( C_4H_{10} \) reacts with 6 moles of \( O_2 \).
- \( O_2 \) needed = \( 0.258 \times 6 \approx 1.55 \text{ moles of } O_2 \)
Difference in oxygen used:
- Difference = \( 3.36 - 1.55 = 1.81 \text{ moles of O}_2 \)
Summary of Results:
41. a) 115g, b) 123g, c) 23.24g of Lithium.
42. a) 107g, b) 1400g, c) 2.87g.
43. a) 98.87g, b) 37.46g.
44. a) 3216.01g, b) 151.33g.
45. a) 7.40g, b) 0.38g, c) Mass decrease due to hydrogen gas escaping.
46. Difference in oxygen consumption between complete and incomplete combustion: 1.81 moles of O2.