To answer this question, we need to employ stoichiometry and the concept of molar masses in chemistry. Here is the step-by-step solution:
1. Identify the Molar Masses:
- The molar mass of [tex]\( \text{AlCl}_3 \)[/tex] (aluminum chloride) is approximately 133.34 g/mol.
- The molar mass of [tex]\( \text{H}_2 \)[/tex] (hydrogen) is approximately 2.02 g/mol.
2. Calculate the Moles of [tex]\( \text{AlCl}_3 \)[/tex] Produced:
- Given that 129 grams of [tex]\( \text{AlCl}_3 \)[/tex] is produced, we need to calculate the moles of [tex]\( \text{AlCl}_3 \)[/tex].
[tex]\[
\text{moles of } \text{AlCl}_3 = \frac{\text{mass of } \text{AlCl}_3}{\text{molar mass of } \text{AlCl}_3} = \frac{129 \text{ g}}{133.34 \text{ g/mol}} \approx 0.967 \text{ moles}
\][/tex]
3. Use Stoichiometry to Find the Moles of [tex]\( \text{H}_2 \)[/tex] Produced:
- According to the balanced chemical equation, 2 moles of [tex]\( \text{AlCl}_3 \)[/tex] are produced along with 3 moles of [tex]\( \text{H}_2 \)[/tex].
- Therefore, the ratio is [tex]\(\frac{3 \text{ moles of } \text{H}_2}{2 \text{ moles of } \text{AlCl}_3}\)[/tex].
- Calculate the moles of [tex]\( \text{H}_2 \)[/tex] produced:
[tex]\[
\text{moles of } \text{H}_2 = \left(\frac{3 \text{ moles of } \text{H}_2}{2 \text{ moles of } \text{AlCl}_3}\right) \times 0.967 \text{ moles} = 1.451 \text{ moles}
\][/tex]
4. Calculate the Mass of [tex]\( \text{H}_2 \)[/tex] Produced:
- Finally, convert the moles of [tex]\( \text{H}_2 \)[/tex] to grams using its molar mass.
[tex]\[
\text{mass of } \text{H}_2 = \text{moles of } \text{H}_2 \times \text{molar mass of } \text{H}_2 = 1.451 \text{ moles} \times 2.02 \text{ g/mol} = 2.93 \text{ grams}
\][/tex]
Thus, the mass of [tex]\( \text{H}_2 \)[/tex] produced when 129 grams of [tex]\( \text{AlCl}_3 \)[/tex] is formed is approximately 2.93 grams.
The correct answer is:
B. 2.92
