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To determine how many grams of [tex]\( \text{Cl}_2 \)[/tex] are consumed to produce 12.0 grams of [tex]\( \text{KCl} \)[/tex], follow these steps:
1. Calculate the molar mass of [tex]\( \text{KCl} \)[/tex] and [tex]\( \text{Cl}_2 \)[/tex]:
- The molar mass of potassium chloride ([tex]\( \text{KCl} \)[/tex]) is given:
[tex]\[ \text{Molar mass of KCl} = 74.55 \, \text{g/mol} \][/tex]
- The molar mass of chlorine gas ([tex]\( \text{Cl}_2 \)[/tex]) is given:
[tex]\[ \text{Molar mass of Cl}_2 = 70.90 \, \text{g/mol} \][/tex]
2. Calculate the moles of [tex]\( \text{KCl} \)[/tex] produced:
- Using the mass of [tex]\( \text{KCl} \)[/tex] produced (12.0 grams):
[tex]\[ \text{Moles of KCl} = \frac{\text{Mass of KCl}}{\text{Molar mass of KCl}} = \frac{12.0 \, \text{g}}{74.55 \, \text{g/mol}} \approx 0.16097 \, \text{mol} \][/tex]
3. Determine the moles of [tex]\( \text{Cl}_2 \)[/tex] required:
- According to the stoichiometry of the reaction [tex]\( 2 \, \text{K} + \text{Cl}_2 \rightarrow 2 \, \text{KCl} \)[/tex]:
- 2 moles of [tex]\( \text{KCl} \)[/tex] are produced from 1 mole of [tex]\( \text{Cl}_2 \)[/tex]
- The moles of [tex]\( \text{Cl}_2 \)[/tex] required to produce the given moles of [tex]\( \text{KCl} \)[/tex]:
[tex]\[ \text{Moles of Cl}_2 = \frac{\text{Moles of KCl}}{2} = \frac{0.16097 \, \text{mol}}{2} \approx 0.08048 \, \text{mol} \][/tex]
4. Calculate the mass of [tex]\( \text{Cl}_2 \)[/tex] consumed:
- Using the moles of [tex]\( \text{Cl}_2 \)[/tex] and the molar mass of [tex]\( \text{Cl}_2 \)[/tex]:
[tex]\[ \text{Mass of Cl}_2 = \text{Moles of Cl}_2 \times \text{Molar mass of Cl}_2 = 0.08048 \, \text{mol} \times 70.90 \, \text{g/mol} \approx 5.706 \, \text{g} \][/tex]
Therefore, to produce 12.0 grams of [tex]\( \text{KCl} \)[/tex], approximately 5.706 grams of [tex]\( \text{Cl}_2 \)[/tex] are consumed.
1. Calculate the molar mass of [tex]\( \text{KCl} \)[/tex] and [tex]\( \text{Cl}_2 \)[/tex]:
- The molar mass of potassium chloride ([tex]\( \text{KCl} \)[/tex]) is given:
[tex]\[ \text{Molar mass of KCl} = 74.55 \, \text{g/mol} \][/tex]
- The molar mass of chlorine gas ([tex]\( \text{Cl}_2 \)[/tex]) is given:
[tex]\[ \text{Molar mass of Cl}_2 = 70.90 \, \text{g/mol} \][/tex]
2. Calculate the moles of [tex]\( \text{KCl} \)[/tex] produced:
- Using the mass of [tex]\( \text{KCl} \)[/tex] produced (12.0 grams):
[tex]\[ \text{Moles of KCl} = \frac{\text{Mass of KCl}}{\text{Molar mass of KCl}} = \frac{12.0 \, \text{g}}{74.55 \, \text{g/mol}} \approx 0.16097 \, \text{mol} \][/tex]
3. Determine the moles of [tex]\( \text{Cl}_2 \)[/tex] required:
- According to the stoichiometry of the reaction [tex]\( 2 \, \text{K} + \text{Cl}_2 \rightarrow 2 \, \text{KCl} \)[/tex]:
- 2 moles of [tex]\( \text{KCl} \)[/tex] are produced from 1 mole of [tex]\( \text{Cl}_2 \)[/tex]
- The moles of [tex]\( \text{Cl}_2 \)[/tex] required to produce the given moles of [tex]\( \text{KCl} \)[/tex]:
[tex]\[ \text{Moles of Cl}_2 = \frac{\text{Moles of KCl}}{2} = \frac{0.16097 \, \text{mol}}{2} \approx 0.08048 \, \text{mol} \][/tex]
4. Calculate the mass of [tex]\( \text{Cl}_2 \)[/tex] consumed:
- Using the moles of [tex]\( \text{Cl}_2 \)[/tex] and the molar mass of [tex]\( \text{Cl}_2 \)[/tex]:
[tex]\[ \text{Mass of Cl}_2 = \text{Moles of Cl}_2 \times \text{Molar mass of Cl}_2 = 0.08048 \, \text{mol} \times 70.90 \, \text{g/mol} \approx 5.706 \, \text{g} \][/tex]
Therefore, to produce 12.0 grams of [tex]\( \text{KCl} \)[/tex], approximately 5.706 grams of [tex]\( \text{Cl}_2 \)[/tex] are consumed.
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