Answer :
To find out how many grams of AlCl₃ are produced when 30.0 g of H₂ are produced, we'll go through the process of stoichiometry. Here are the steps:
1. Balance the Chemical Equation:
[tex]\[ 2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2 \][/tex]
2. Understand the Stoichiometry:
According to the balanced equation, 3 moles of H₂ are produced along with 2 moles of AlCl₃.
3. Calculate the Moles of H₂ Produced:
First, find the moles of H₂ produced from the given mass.
- Molar mass of H₂ is approximately 2.02 g/mol.
- Moles of H₂ = [tex]\(\frac{\text{mass of } H_2}{\text{molar mass of } H_2}\)[/tex]
- Moles of H₂ = [tex]\(\frac{30.0 \text{ g}}{2.02 \text{ g/mol}}\)[/tex]
- Moles of H₂ ≈ 14.85 moles
4. Use Stoichiometry to Find Moles of AlCl₃ Produced:
Based on the balanced equation, 3 moles of H₂ are produced alongside 2 moles of AlCl₃.
- Moles of AlCl₃ = [tex]\(\frac{2}{3} \times \text{moles of H₂}\)[/tex]
- Moles of AlCl₃ = [tex]\(\frac{2}{3} \times 14.85\)[/tex]
- Moles of AlCl₃ ≈ 9.90 moles
5. Calculate the Mass of AlCl₃ Produced:
Now calculate the mass of AlCl₃ using its molar mass (133.33 g/mol).
- Mass of AlCl₃ = [tex]\(\text{moles of AlCl₃} \times \text{molar mass of AlCl₃}\)[/tex]
- Mass of AlCl₃ = [tex]\(9.90 \times 133.33 \text{ g/mol}\)[/tex]
- Mass of AlCl₃ ≈ 1320 g
Therefore, the amount of AlCl₃ produced when 30.0 g of H₂ are produced is approximately 1320 g, which corresponds to option 4 in the provided choices.
1. Balance the Chemical Equation:
[tex]\[ 2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2 \][/tex]
2. Understand the Stoichiometry:
According to the balanced equation, 3 moles of H₂ are produced along with 2 moles of AlCl₃.
3. Calculate the Moles of H₂ Produced:
First, find the moles of H₂ produced from the given mass.
- Molar mass of H₂ is approximately 2.02 g/mol.
- Moles of H₂ = [tex]\(\frac{\text{mass of } H_2}{\text{molar mass of } H_2}\)[/tex]
- Moles of H₂ = [tex]\(\frac{30.0 \text{ g}}{2.02 \text{ g/mol}}\)[/tex]
- Moles of H₂ ≈ 14.85 moles
4. Use Stoichiometry to Find Moles of AlCl₃ Produced:
Based on the balanced equation, 3 moles of H₂ are produced alongside 2 moles of AlCl₃.
- Moles of AlCl₃ = [tex]\(\frac{2}{3} \times \text{moles of H₂}\)[/tex]
- Moles of AlCl₃ = [tex]\(\frac{2}{3} \times 14.85\)[/tex]
- Moles of AlCl₃ ≈ 9.90 moles
5. Calculate the Mass of AlCl₃ Produced:
Now calculate the mass of AlCl₃ using its molar mass (133.33 g/mol).
- Mass of AlCl₃ = [tex]\(\text{moles of AlCl₃} \times \text{molar mass of AlCl₃}\)[/tex]
- Mass of AlCl₃ = [tex]\(9.90 \times 133.33 \text{ g/mol}\)[/tex]
- Mass of AlCl₃ ≈ 1320 g
Therefore, the amount of AlCl₃ produced when 30.0 g of H₂ are produced is approximately 1320 g, which corresponds to option 4 in the provided choices.