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Sagot :
To solve this problem, we need to follow a step-by-step approach, using stoichiometry and the given data:
1. Identify the given information and what is being asked:
- Given: 129 grams of [tex]\( AlCl_3 \)[/tex] produced.
- Task: Find the mass of [tex]\( H_2 \)[/tex] produced in grams.
2. Calculate the molar masses of the relevant compounds:
- Molar mass of [tex]\( AlCl_3 \)[/tex]:
- [tex]\( Al \)[/tex]: [tex]\( 26.98 \, g/mol \)[/tex]
- [tex]\( Cl \)[/tex]: [tex]\( 35.45 \, g/mol \)[/tex]
- [tex]\( AlCl_3 = 26.98 + 3 \times 35.45 = 133.34 \, g/mol \)[/tex]
- Molar mass of [tex]\( H_2 \)[/tex]:
- [tex]\( H \)[/tex]: [tex]\( 1.01 \, g/mol \)[/tex]
- [tex]\( H_2 = 2 \times 1.01 = 2.02 \, g/mol \)[/tex]
3. Determine the moles of [tex]\( AlCl_3 \)[/tex] produced using its molar mass:
- [tex]\( \text{moles of } AlCl_3 = \frac{\text{mass of } AlCl_3}{\text{molar mass of } AlCl_3} \)[/tex]
- [tex]\( \text{moles of } AlCl_3 = \frac{129 \, g}{133.34 \, g/mol} \)[/tex]
- [tex]\( \text{moles of } AlCl_3 \approx 0.96758 \)[/tex]
4. Use the balanced chemical equation to relate the moles of [tex]\( AlCl_3 \)[/tex] to the moles of [tex]\( H_2 \)[/tex]:
[tex]\[ 2 \, AlCl_3 \leftrightarrow 3 \, H_2 \][/tex]
- From the equation, 2 moles of [tex]\( AlCl_3 \)[/tex] produce 3 moles of [tex]\( H_2 \)[/tex].
- Therefore, 1 mole of [tex]\( AlCl_3 \)[/tex] would produce [tex]\( \frac{3}{2} \)[/tex] moles of [tex]\( H_2 \)[/tex].
- [tex]\( \text{moles of } H_2 = \left( \frac{3}{2} \right) \times \text{moles of } AlCl_3 \)[/tex]
- [tex]\( \text{moles of } H_2 = \left( \frac{3}{2} \right) \times 0.96758 \)[/tex]
- [tex]\( \text{moles of } H_2 \approx 1.45137 \)[/tex]
5. Convert the moles of [tex]\( H_2 \)[/tex] to grams using the molar mass of [tex]\( H_2 \)[/tex]:
- [tex]\( \text{mass of } H_2 = \text{moles of } H_2 \times \text{molar mass of } H_2 \)[/tex]
- [tex]\( \text{mass of } H_2 = 1.45137 \, moles \times 2.02 \, g/mol \)[/tex]
- [tex]\( \text{mass of } H_2 \approx 2.93138 \, g \)[/tex]
6. Round the final result for clarity:
- [tex]\( \text{mass of } H_2 \approx 2.92 \, g \)[/tex]
Thus, the correct answer is:
B. 2.92
1. Identify the given information and what is being asked:
- Given: 129 grams of [tex]\( AlCl_3 \)[/tex] produced.
- Task: Find the mass of [tex]\( H_2 \)[/tex] produced in grams.
2. Calculate the molar masses of the relevant compounds:
- Molar mass of [tex]\( AlCl_3 \)[/tex]:
- [tex]\( Al \)[/tex]: [tex]\( 26.98 \, g/mol \)[/tex]
- [tex]\( Cl \)[/tex]: [tex]\( 35.45 \, g/mol \)[/tex]
- [tex]\( AlCl_3 = 26.98 + 3 \times 35.45 = 133.34 \, g/mol \)[/tex]
- Molar mass of [tex]\( H_2 \)[/tex]:
- [tex]\( H \)[/tex]: [tex]\( 1.01 \, g/mol \)[/tex]
- [tex]\( H_2 = 2 \times 1.01 = 2.02 \, g/mol \)[/tex]
3. Determine the moles of [tex]\( AlCl_3 \)[/tex] produced using its molar mass:
- [tex]\( \text{moles of } AlCl_3 = \frac{\text{mass of } AlCl_3}{\text{molar mass of } AlCl_3} \)[/tex]
- [tex]\( \text{moles of } AlCl_3 = \frac{129 \, g}{133.34 \, g/mol} \)[/tex]
- [tex]\( \text{moles of } AlCl_3 \approx 0.96758 \)[/tex]
4. Use the balanced chemical equation to relate the moles of [tex]\( AlCl_3 \)[/tex] to the moles of [tex]\( H_2 \)[/tex]:
[tex]\[ 2 \, AlCl_3 \leftrightarrow 3 \, H_2 \][/tex]
- From the equation, 2 moles of [tex]\( AlCl_3 \)[/tex] produce 3 moles of [tex]\( H_2 \)[/tex].
- Therefore, 1 mole of [tex]\( AlCl_3 \)[/tex] would produce [tex]\( \frac{3}{2} \)[/tex] moles of [tex]\( H_2 \)[/tex].
- [tex]\( \text{moles of } H_2 = \left( \frac{3}{2} \right) \times \text{moles of } AlCl_3 \)[/tex]
- [tex]\( \text{moles of } H_2 = \left( \frac{3}{2} \right) \times 0.96758 \)[/tex]
- [tex]\( \text{moles of } H_2 \approx 1.45137 \)[/tex]
5. Convert the moles of [tex]\( H_2 \)[/tex] to grams using the molar mass of [tex]\( H_2 \)[/tex]:
- [tex]\( \text{mass of } H_2 = \text{moles of } H_2 \times \text{molar mass of } H_2 \)[/tex]
- [tex]\( \text{mass of } H_2 = 1.45137 \, moles \times 2.02 \, g/mol \)[/tex]
- [tex]\( \text{mass of } H_2 \approx 2.93138 \, g \)[/tex]
6. Round the final result for clarity:
- [tex]\( \text{mass of } H_2 \approx 2.92 \, g \)[/tex]
Thus, the correct answer is:
B. 2.92
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