To solve these questions, we will apply the principles of stoichiometry based on the given chemical equations. This includes using molar masses, determining mole ratios, and calculating the amounts of reactants or products needed or produced.
41. Lithium and Fluorine Reaction
The balanced equation is: \[ 2 \text{Li} + \text{F}_2 \rightarrow 2 \text{LiF} \]
a. Calculate the mass of Li needed to produce 115g of LiF.
- Molar mass of Li = 6.94 g/mol
- Molar mass of LiF = 6.94 g/mol (Li) + 19.00 g/mol (F) = 25.94 g/mol
- Moles of LiF produced = \( \frac{115 , \text{g}}{25.94 , \text{g/mol}} \approx 4.43 , \text{mol} \)
- According to the reaction, 2 moles of Li produce 2 moles of LiF, so moles of Li needed = moles of LiF = 4.43 mol
- Mass of Li = \( 4.43 , \text{mol} \times 6.94 , \text{g/mol} \approx 30.73 , \text{g} \)
b. Calculate the mass of F2 needed to produce 123g of LiF.
- Moles of LiF produced = \( \frac{123 , \text{g}}{25.94 , \text{g/mol}} \approx 4.74 , \text{mol} \)
- Moles of F2 needed = \( \frac{1 , \text{mol F}_2}{2 , \text{mol LiF}} \times 4.74 , \text{mol LiF} = 2.37 , \text{mol F}_2 \)
- Mass of F2 = \( 2.37 , \text{mol} \times 38.00 , \text{g/mol} \approx 90.06 , \text{g} \)
c. What mass of Li was used if 87g of LiF is produced?
- Moles of LiF = \( \frac{87 , \text{g}}{25.94 , \text{g/mol}} \approx 3.35 , \text{mol} \)
- Moles of Li = 3.35 mol
- Mass of Li = \( 3.35 , \text{mol} \times 6.94 , \text{g/mol} \approx 23.25 , \text{g} \)
42. Calcium and Oxygen Reaction
The balanced equation is: \[ 2 \text{Ca} + \text{O}_2 \rightarrow 2 \text{CaO} \]
a. Calculate the mass of Ca needed to produce 150g of CaO.
- Molar mass of CaO = 40.08 g/mol + 16.00 g/mol = 56.08 g/mol
- Moles of CaO = \( \frac{150 , \text{g}}{56.08 , \text{g/mol}} \approx 2.67 , \text{mol} \)
- Moles of Ca needed = 2.67 mol
- Mass of Ca = \( 2.67 , \text{mol} \times 40.08 , \text{g/mol} \approx 107.05 , \text{g} \)
b. Calculate mass of CaO produced from 1 kg of Ca.
- Moles of Ca = \( \frac{1000 , \text{g}}{40.08 , \text{g/mol}} \approx 24.93 , \text{mol} \)
- Moles of CaO produced = 24.93 mol
- Mass of CaO = \( 24.93 , \text{mol} \times 56.08 , \text{g/mol} \approx 1397.57 , \text{g} \)
c. Calculate mass of CaO from 0.82 g of O2.
- Moles of O2 = \( \frac{0.82 , \text{g}}{32.00 , \text{g/mol}} \approx 0.0256 , \text{mol} \)
- Moles of CaO produced = \( 2 \times 0.0256 , \text{mol} \) = 0.0512 mol
- Mass of CaO = \( 0.0512 , \text{mol} \times 56.08 , \text{g/mol} \approx 2.87 , \text{g} \)
d. Calculate number of O2 molecules in 0.82 g.
- Moles of O2 = 0.0256 mol
- Number of molecules = \( 0.0256 , \text{mol} \times 6.02 \times 10^{23} , \text{molecules/mol} \approx 1.54 \times 10^{22} \)
e. Explain why the mass appears to increase. The mass appears to increase due to the formation of a solid product (calcium oxide) from gaseous reactants (calcium and oxygen), which can also include errors in measurement or considerations of the conservation of mass.
43. Phosphorus and Oxygen Reaction
The balanced equation is: \[ \text{P}_4 + 5 \text{O}_2 \rightarrow 2 \text{P}_2\text{O}_5 \]
a. Calculate the product formed from 43 g of phosphorous.
- Molar mass of P4 = 4 × 30.97 g/mol = 123.88 g/mol
- Moles of P4 = \( \frac{43, \text{g}}{123.88 , \text{g/mol}} \approx 0.348 , \text{mol} \)
- From the reaction, 1 mole of P4 produces 2 moles of P2O5.
- Moles of P2O5 = \( 0.348 , \text{mol} \times 2 = 0.696 , \text{mol} \)
- Molar mass of P2O5 = 2 × 30.97 + 5 × 16 = 141.94 g/mol
- Mass of P2O5 = \( 0.696 , \text{mol} \times 141.94 , \text{g/mol} \approx 98.90 , \text{g} \)
b. Calculate product formed with 21 g of oxygen.
- Moles of O2 = \( \frac{21 , \text{g}}{32.00 , \text{g/mol}} \approx 0.656 , \text{mol} \)
- Moles of P2O5 = \( \frac{2}{5} \times 0.656 , \text{mol} = 0.263 , \text{mol} \)
- Mass of P2O5 = \( 0.263 , \text{mol} \times 141.94 , \text{g/mol} \approx 37.40 , \text{g} \)
c. Calculate mass of O2 needed for 90g of P2O5.
- Moles of P2O5 = \( \frac{90 , \text{g}}{141.94 , \text{g/mol}} \approx 0.633 , \text{mol} \)
- Moles of O2 needed = \( \frac{5}{2} \times 0.633 = 1.583 , \text{mol} \)
- Mass of O2 = \( 1.583 , \text{mol} \times 32.00 , \text{g/mol} \approx 50.576 , \text{g} \)
44. Aluminium and Chlorine Reaction
The balanced equation is: \[ 4 \text{Al} + 3 \text{Cl}_2 \rightarrow 2 \text{Al}_2\text{Cl}_3 \]
a. If 1.3 kg of Al is used, how much Al2Cl3 is produced?
- Molar mass of Al = 26.98 g/mol
- Moles of Al = \( \frac{1300 , \text{g}}{26.98 , \text{g/mol}} \approx 48.31 , \text{mol} \)
- According to the stoichiometry, 4 moles of Al produce 2 moles of Al2Cl3.
- Moles of Al2Cl3 produced = \( \frac{2}{4} \times 48.31 \approx 24.15 , \text{mol} \)
- Molar mass of Al2Cl3 = 2 × 26.98 + 3 × 35.45 = 133.33 g/mol
- Mass of Al2Cl3 = \( 24.15 , \text{mol} \times 133.33 , \text{g/mol} \approx 3217.15 , \text{g} \)
b. Mass of chlorine required for 77g of Al.
- Moles of Al = \( \frac{77 , \text{g}}{26.98 , \text{g/mol}} \approx 2.86 , \text{mol} \)
- Moles of Cl2 needed = \( \frac{3}{4} \times 2.86 \approx 2.145 , \text{mol} \)
- Mass of Cl2 = \( 2.145 , \text{mol} \times 70.91 , \text{g/mol} \approx 152.08 , \text{g} \)
45. Potassium and Water Reaction
The balanced equation is: \[ 2 \text{K} + 2 \text{H}_2\text{O} \rightarrow 2 \text{KOH} + \text{H}_2 \]
a. How much hydrogen is produced from 132g of water?
- Molar mass of water = 18.02 g/mol
- Moles of H2O = \( \frac{132 , \text{g}}{18.02 , \text{g/mol}} \approx 7.33 , \text{mol} \)
- From the equation, 2 moles of H2O produce 1 mole of H2.
- Moles of H2 produced = \( \frac{1}{2} \times 7.33 = 3.665 , \text{mol} \)
- Mass of H2 = \( 3.665 , \text{mol} \times 2.02 , \text{g/mol} \approx 7.39 , \text{g} \)
b. How much potassium is required to produce 0.55g of KOH?
- Molar mass of KOH = 39.10 + 16.00 + 1.01 = 56.11 g/mol
- Moles of KOH = \( \frac{0.55 , \text{g}}{56.11 , \text{g/mol}} \approx 0.00979 , \text{mol} \)
- Moles of K needed = \( 0.00979 , \text{mol} \)
- Mass of K = \( 0.00979 , \text{mol} \times 39.10 , \text{g/mol} \approx 0.384 , \text{g} \)
c. Suggest why the mass appears to decrease. The mass decreases because hydrogen gas is released into the atmosphere, not being captured during the reaction, making the final mass of products lighter than the initial reactants.
46. Challenge: Butane Combustion
Complete combustion reaction: \[ \text{C}4\text{H}{10} + 6 \text{O}_2 \rightarrow 4 \text{CO}_2 + 5 \text{H}_2\text{O} \]
- Molar mass of C4H10 = \( (4 \times 12.01) + (10 \times 1.008) = 58.12 , \text{g/mol} \)
- Moles of C4H10 = \( \frac{15 , \text{g}}{58.12 , \text{g/mol}} \approx 0.258 , \text{mol} \)
- According to the reaction, 1 mole of butane needs 6 moles of O2.
- Moles of O2 required for complete combustion = \( 0.258 \times 6 = 1.548 , \text{mol} \)
Incomplete combustion reaction: \[ \text{C}4\text{H}{10} + 2 \text{O}_2 \rightarrow 4 \text{CO} + 5 \text{H}_2\text{O} \]
- Moles of O2 required for incomplete combustion = \( 0.258 \times 2 = 0.516 , \text{mol} \)
Comparison of oxygen required:
- Complete combustion: 1.548 mol O2
- Incomplete combustion: 0.516 mol O2
The difference is:
- Oxygen needed = \( 1.548 - 0.516 = 1.032 , \text{mol} \)
These calculations demonstrate how the amount of oxygen differs between complete and incomplete combustion reactions of butane.