Let's work through the questions one by one:
41. Reaction: \( 2\text{Li} + \text{F}_2 \rightarrow 2\text{LiF} \)
a. Show that 31 g of Lithium produces approximately 115 g of lithium fluoride.
Molar mass of Lithium (Li):
- \( \text{Li} = 6.94 , \text{g/mol} \)
Calculating moles of Lithium used: \[ \text{moles of Li} = \frac{31 , \text{g}}{6.94 , \text{g/mol}} \approx 4.47 , \text{mol} \]
From the balanced equation, \( 2 , \text{mol of Li} \) produces \( 2 , \text{mol of LiF} \).
Therefore, \( 4.47 , \text{mol of Li} \) produces \( 4.47 , \text{mol of LiF} \).
Molar mass of Lithium Fluoride (LiF):
- \( \text{LiF} = 6.94 , \text{g/mol} + 19.00 , \text{g/mol} = 25.94 , \text{g/mol} \)
Calculating mass of LiF produced: \[ \text{mass of LiF} = 4.47 , \text{mol} \times 25.94 , \text{g/mol} \approx 115.0 , \text{g} \]
Thus, 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.
Molar mass of Fluorine (F2):
- \( \text{F}_2 = 19.00 , \text{g/mol} \times 2 = 38.00 , \text{g/mol} \)
Calculating moles of Fluorine used: \[ \text{moles of F2} = \frac{90 , \text{g}}{38.00 , \text{g/mol}} \approx 2.37 , \text{mol} \]
From the balanced equation, \( 1 , \text{mol of F2} \) produces \( 2 , \text{mol of LiF} \).
So, \( 2.37 , \text{mol of F2} \) produces:
\[
2 \times 2.37 , \text{mol} = 4.74 , \text{mol of LiF}
\]
Calculating mass of LiF produced: \[ \text{mass of LiF} = 4.74 , \text{mol} \times 25.94 , \text{g/mol} \approx 122.99 , \text{g} \approx 123 , \text{g} \]
Thus, 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?
Calculating moles of LiF produced: \[ \text{moles of LiF} = \frac{87 , \text{g}}{25.94 , \text{g/mol}} \approx 3.36 , \text{mol} \]
From the balanced equation, \( 2 , \text{mol of LiF} \) requires \( 2 , \text{mol of Li} \).
So, \( 3.36 , \text{mol of LiF} \) requires:
\[
3.36 , \text{mol of Li}
\]
Calculating mass of lithium used: \[ \text{mass of Li} = 3.36 , \text{mol} \times 6.94 , \text{g/mol} \approx 23.30 , \text{g} \]
42. Reaction: \( 2\text{Ca} + \text{O}_2 \rightarrow 2\text{CaO} \)
a. Show that approximately 107 g of calcium is needed to produce 150 g of calcium oxide.
Molar mass of Calcium Oxide (CaO):
- \( \text{CaO} = 40.08 , \text{g/mol} + 16.00 , \text{g/mol} = 56.08 , \text{g/mol} \)
Calculate moles of CaO: \[ \text{moles of CaO} = \frac{150 , \text{g}}{56.08 , \text{g/mol}} \approx 2.67 , \text{mol} \]
From the balanced equation:
\( 2 , \text{mol of Ca} \) produces \( 2 , \text{mol of CaO} \)
So, \( 2.67 , \text{mol of CaO} \) requires:
\[
2.67 , \text{mol of Ca}
\]
Calculating mass of Ca: \[ \text{mass of Ca} = 2.67 , \text{mol} \times 40.08 , \text{g/mol} \approx 107.0 , \text{g} \]
Thus, 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.
\[ \text{mass of Ca} = 1000 , \text{g} \] Calculating moles of Ca: \[ \text{moles of Ca} = \frac{1000 , \text{g}}{40.08 , \text{g/mol}} \approx 24.93 , \text{mol} \]
Calculating moles of CaO produced: From the balanced equation, \( 2 , \text{mol of Ca} \) produces \( 2 , \text{mol of CaO} \) Thus \( 24.93 , \text{mol of Ca} \) produces \( 24.93 , \text{mol of CaO} \).
Calculating mass of CaO produced: \[ \text{mass of CaO} = 24.93 , \text{mol} \times 56.08 , \text{g/mol} \approx 1399.5 , \text{g} \approx 1400 , \text{g} \]
Thus, the mass of calcium oxide produced from 1 kg of calcium is approximately 1400 g.
c. Calculate the mass of calcium oxide formed from 0.82 g of oxygen.
Molar mass of O2:
- \( \text{O}_2 = 32.00 , \text{g/mol} \)
Calculating moles of O2: \[ \text{moles of O2} = \frac{0.82 , \text{g}}{32.00 , \text{g/mol}} \approx 0.0256 , \text{mol} \]
From the balanced equation \( 1 , \text{mol of O2} \) produces \( 2 , \text{mol of CaO} \)
So, \( 0.0256 , \text{mol of O2} \) produces:
\[
2 \times 0.0256 , \text{mol} = 0.0512 , \text{mol of CaO}
\]
Calculating mass of CaO produced: \[ \text{mass of CaO} = 0.0512 , \text{mol} \times 56.08 , \text{g/mol} \approx 2.87 , \text{g} \]
d. How many molecules of oxygen are in 0.82 g of oxygen?
Using Avogadro's number \( (6.022 \times 10^{23} , \text{molecules/mol}) \): \[ \text{molecules of O2} = 0.0256 , \text{mol} \times 6.022 \times 10^{23} , \text{molecules/mol} \approx 1.54 \times 10^{22} , \text{molecules} \]
e. Explain why the mass of this reaction appears to increase.
This reaction mass does not actually increase; it appears that way because of the gas reactions. Since gases (specifically O2 in this case) can occupy a significant volume when produced, if the system does not measure gas accurately, it may seem like mass increases. Also, if the calcium oxide formed appears to result in higher mass, it might come from the formation of solid products compared to gaseous reactants.
43. \( \text{P}_4 + 5\text{O}_2 \rightarrow 2\text{P}_2\text{O}_5 \)
a. Show that the amount of product formed from 43g of phosphorus is just less than 100g.
Molar mass of Phosphorus (P4):
- \( \text{P}_4 = 4 \times 30.97 , \text{g/mol} = 123.88 , \text{g/mol} \)
Calculating moles of P4: \[ \text{moles of P4} = \frac{43 , \text{g}}{123.88 , \text{g/mol}} \approx 0.347 , \text{mol} \]
From the balanced equation, \( 1 , \text{mol of P4} \) produces \( 2 , \text{mol of P2O5} \).
Thus \( 0.347 , \text{mol of P4} \) produces:
\[
2 \times 0.347 , \text{mol} = 0.694 , \text{mol of P2O5}
\]
Molar mass of P2O5:
- \( \text{P}_2\text{O}_5 = 2 \times 30.97 + 5 \times 16.00 = 142.97 , \text{g/mol} \)
Calculating mass of P2O5 produced: \[ \text{mass of P2O5} = 0.694 , \text{mol} \times 142.97 , \text{g/mol} \approx 99.305 , \text{g} \approx 99 , \text{g} \]
Thus, 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?
Molar mass of O2:
- \( \text{O}_2 = 32.00 , \text{g/mol} \)
Calculating moles of O2: \[ \text{moles of O2} = \frac{21 , \text{g}}{32.00 , \text{g/mol}} \approx 0.656 , \text{mol} \]
From the balanced equation, \( 5 , \text{mol of O2} \) produces \( 2 , \text{mol of P2O5} \).
Thus \( 0.656 , \text{mol of O2} \) produces:
\[
\frac{2}{5} \times 0.656 , \text{mol} = 0.2624 , \text{mol of P2O5}
\]
Calculating mass of P2O5 produced: \[ \text{mass of P2O5} = 0.2624 , \text{mol} \times 142.97 , \text{g/mol} \approx 37.5 , \text{g} \]
c. How much oxygen would be needed for 90g of product?
Calculating moles of P2O5 produced: \[ \text{moles of P2O5} = \frac{90 , \text{g}}{142.97 , \text{g/mol}} \approx 0.629 , \text{mol} \]
From the balanced equation, \( 2 , \text{mol of P2O5} \) requires \( 5 , \text{mol of O2} \).
Thus \( 0.629 , \text{mol of P2O5} \) requires:
\[
\frac{5}{2} \times 0.629 , \text{mol} = 1.573 , \text{mol of O2}
\]
Calculating mass of O2 needed: \[ \text{mass of O2} = 1.573 , \text{mol} \times 32.00 , \text{g/mol} \approx 50.0 , \text{g} \]
44. \( 4\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{Al}_2\text{Cl}_3 \)
a. 1.3 kg of aluminium is used. How much aluminium chloride is produced?
Molar mass of Aluminium (Al):
- \( \text{Al} = 26.98 , \text{g/mol} \)
Calculating moles of Al: \[ \text{mass of Al} = 1300 , \text{g} \] \[ \text{moles of Al} = \frac{1300 , \text{g}}{26.98 , \text{g/mol}} \approx 48.22 , \text{mol} \]
From the balanced equation, \( 4 , \text{mol of Al} \) produces \( 2 , \text{mol of Al}_2\text{Cl}_3 \).
Thus \( 48.22 , \text{mol of Al} \) produces:
\[
\frac{2}{4} \times 48.22 , \text{mol} = 24.11 , \text{mol of Al}_2\text{Cl}_3
\]
Molar mass of Al2Cl3:
- \( \text{Al}_2\text{Cl}_3 = 2 \times 26.98 , \text{g/mol} + 3 \times 35.45 , \text{g/mol} = 133.33 , \text{g/mol} \)
Calculating mass of Al2Cl3 produced: \[ \text{mass of Al}_2\text{Cl}_3 = 24.11 , \text{mol} \times 133.33 , \text{g/mol} \approx 3215.59 , \text{g} \approx 3.22 , \text{kg} \]
b. How much chlorine is required to react with 77g of aluminium?
Calculating moles of Al: \[ \text{moles of Al} = \frac{77 , \text{g}}{26.98 , \text{g/mol}} \approx 2.86 , \text{mol} \]
From the balanced equation, \( 4 , \text{mol of Al} \) requires \( 3 , \text{mol of Cl2} \)
Thus \( 2.86 , \text{mol of Al} \) requires:
\[
\frac{3}{4} \times 2.86 , \text{mol} = 2.14 , \text{mol of Cl2}
\]
Calculating mass of Cl2 needed: \[ \text{mass of Cl2} = 2.14 , \text{mol} \times 70.91 , \text{g/mol} \approx 151.63 , \text{g} \]
45. \( 2\text{K} + 2\text{H}_2\text{O} \rightarrow 2\text{KOH} + \text{H}_2 \)
a. How much hydrogen is produced from a reaction using 132g of water?
Molar mass of water (H2O):
- \( \text{H}_2\text{O} = 18.02 , \text{g/mol} \)
Calculating moles of water: \[ \text{moles of H2O} = \frac{132 , \text{g}}{18.02 , \text{g/mol}} \approx 7.33 , \text{mol} \]
From the balanced equation, \( 2 , \text{mol of H2O} \) produces \( 1 , \text{mol of H2} \).
Thus \( 7.33 , \text{mol of H2O} \) produces:
\[
\frac{1}{2} \times 7.33 , \text{mol} = 3.67 , \text{mol of H2}
\]
Calculating mass of H2 produced: \[ \text{mass of H2} = 3.67 , \text{mol} \times 2.02 , \text{g/mol} \approx 7.40 , \text{g} \]
b. How much potassium is required to produce 0.55g of potassium hydroxide?
Molar mass of KOH:
- \( \text{KOH} = 39.10 , \text{g/mol} + 16.00 , \text{g/mol} + 1.01 , \text{g/mol} = 56.11 , \text{g/mol} \)
Calculating moles of KOH: \[ \text{moles of KOH} = \frac{0.55 , \text{g}}{56.11 , \text{g/mol}} \approx 0.0098 , \text{mol} \]
From the balanced equation, \( 2 , \text{mol of K} \) produces \( 2 , \text{mol of KOH} \).
Thus \( 0.0098 , \text{mol of KOH} \) requires:
\[
0.0098 , \text{mol of K}
\]
Calculating mass of K required: \[ \text{mass of K} = 0.0098 , \text{mol} \times 39.10 , \text{g/mol} \approx 0.384 , \text{g} \]
c. Suggest why the mass of this reaction appears to decrease.
In reactions of metals with water, the hydrogen gas produced escapes into the atmosphere. The mass of the reactants is greater than the mass of the products because the hydrogen gas has escaped, making it appear that mass is lost.
46. Challenge: \( \text{C}4\text{H}{10} \) combustion
Complete combustion reaction: \[ \text{C}4\text{H}{10} + 13/2 , \text{O}_2 \rightarrow 4 , \text{CO}_2 + 5 , \text{H}_2\text{O} \]
Incomplete combustion reaction: \[ \text{C}4\text{H}{10} + 9/2 , \text{O}_2 \rightarrow 4 , \text{CO} + 5 , \text{H}_2\text{O} \]
Starting with 15g of butane, let's determine the mass of oxygen used in both reactions.
- Molar mass of butane (C4H10):
- \( 4 \times 12.01 + 10 \times 1.008 = 58.12 , \text{g/mol} \)
Calculating moles of butane: \[ \text{moles of C}4\text{H}{10} = \frac{15 , \text{g}}{58.12 , \text{g/mol}} \approx 0.258 , \text{mol} \]
-
Complete combustion:
- Requires \( \frac{13}{2} \) moles of \( O_2 \): \[ \text{moles of O}_2 = 0.258 \times \frac{13}{2} = 1.677 , \text{mol} \]
Calculating mass of O2: \[ \text{mass of O}_2 = 1.677 , \text{mol} \times 32.00 , \text{g/mol} = 53.66 , \text{g} \]
-
Incomplete combustion:
- Requires \( \frac{9}{2} \) moles of \( O_2 \): \[ \text{moles of O}_2 = 0.258 \times \frac{9}{2} = 1.162 , \text{mol} \]
Calculating mass of O2: \[ \text{mass of O}_2 = 1.162 , \text{mol} \times 32.00 , \text{g/mol} = 37.18 , \text{g} \]
Summary
- Complete combustion of butane uses approximately 53.66g of O2.
- Incomplete combustion of butane uses approximately 37.18g of O2.
The difference in oxygen needed can be attributed to the incomplete combustion producing carbon monoxide, rather than carbon dioxide which requires more oxygen.
Let me know if you need further explanations or calculations!