Sure, let's solve the problem step by step.
### Step 1: Determine moles of AlCl₃
First, we need to find out how many moles of AlCl₃ are generated from 45.2 g of AlCl₃. The molar mass of AlCl₃ is given as 133.33 g/mol.
[tex]\[
\text{Moles of AlCl}_3 = \frac{\text{mass of AlCl}_3}{\text{molar mass of AlCl}_3}
\][/tex]
[tex]\[
\text{Moles of AlCl}_3 = \frac{45.2 \text{ g}}{133.33 \text{ g/mol}} \approx 0.339 \text{ moles of AlCl}_3
\][/tex]
### Step 2: Relate moles of AlCl₃ to moles of HCl
From the balanced chemical equation, we know that:
[tex]\[
2 \text{ AlCl}_3 \text{ is produced by 6 HCl}
\][/tex]
So, for 1 mole of AlCl₃, there are 3 moles of HCl needed (since 2 moles of AlCl₃ correspond to 6 moles of HCl).
[tex]\[
\text{Moles of HCl} = \text{Moles of AlCl}_3 \times 3
\][/tex]
[tex]\[
\text{Moles of HCl} = 0.339 \text{ moles of AlCl}_3 \times 3 = 1.017 \text{ moles of HCl}
\][/tex]
### Step 3: Determine mass of HCl needed
Finally, we need to find out the mass of this required moles of HCl. The molar mass of HCl is given as 36.46 g/mol.
[tex]\[
\text{Mass of HCl} = \text{moles of HCl} \times \text{molar mass of HCl}
\][/tex]
[tex]\[
\text{Mass of HCl} = 1.017 \text{ moles} \times 36.46 \text{ g/mol} \approx 37.08 \text{ g of HCl}
\][/tex]
### Conclusion
Hence, the mass of HCl needed to generate 45.2 g of AlCl₃ is approximately 37.08 g.
