To determine how many grams of [tex]\( H_2 \)[/tex] are produced when 129 grams of [tex]\( AlCl_3 \)[/tex] are formed, follow these detailed steps:
1. Determine the molar masses of the chemicals involved:
- Aluminum chloride ([tex]\( AlCl_3 \)[/tex]):
- Molar mass of [tex]\( Al \)[/tex] (Aluminum) = 26.98 g/mol
- Molar mass of [tex]\( Cl \)[/tex] (Chlorine) = 35.453 g/mol
- Molar mass of [tex]\( AlCl_3 \)[/tex] = 26.98 + (3 × 35.453) = 26.98 + 106.359 = 133.339 g/mol
- Molecular hydrogen ([tex]\( H_2 \)[/tex]):
- Molar mass of [tex]\( H \)[/tex] (Hydrogen) = 1.008 g/mol
- Molar mass of [tex]\( H_2 \)[/tex] = 2 × 1.008 = 2.016 g/mol
2. Calculate the amount (in moles) of [tex]\( AlCl_3 \)[/tex] produced:
[tex]\[
\text{Moles of } AlCl_3 = \frac{\text{Mass of } AlCl_3}{\text{Molar mass of } AlCl_3} = \frac{129 \text{ g}}{133.34 \text{ g/mol}}
\][/tex]
- Moles of [tex]\( AlCl_3 \)[/tex] ≈ 0.967 moles
3. Use the stoichiometric ratio to find the moles of [tex]\( H_2 \)[/tex] produced:
The balanced chemical equation provides the stoichiometry:
[tex]\[
2 Al + 6 HCl \rightarrow 2 AlCl_3 + 3 H_2
\][/tex]
- From this reaction, 2 moles of [tex]\( AlCl_3 \)[/tex] produce 3 moles of [tex]\( H_2 \)[/tex].
- Therefore, the mole ratio of [tex]\( AlCl_3 \)[/tex] to [tex]\( H_2 \)[/tex] is [tex]\( \frac{3}{2} \)[/tex].
Using the mole ratio:
[tex]\[
\text{Moles of } H_2 = \text{Moles of } AlCl_3 \times \frac{3}{2} = 0.967 \text{ moles} \times 1.5 \approx 1.451 \text{ moles}
\][/tex]
4. Convert moles of [tex]\( H_2 \)[/tex] into grams:
[tex]\[
\text{Mass of } H_2 = \text{Moles of } H_2 \times \text{Molar mass of } H_2 = 1.451 \text{ moles} \times 2.016 \text{ g/mol} \approx 2.925 \text{ grams}
\][/tex]
Thus, the mass of [tex]\( H_2 \)[/tex] produced when 129 grams of [tex]\( AlCl_3 \)[/tex] are formed is approximately 2.92 grams. Therefore, the correct answer is:
B. 2.92
