Let's determine the mass of HCl needed to react completely with 12.8 g of aluminium using the given chemical reaction:
[tex]\[ 2 \text{Al} (s) + 6 \text{HCl} (aq) \rightarrow 2 \text{AlCl}_3 (aq) + 3 \text{H}_2 (g) \][/tex]
### Step-by-Step Solution:
1. Determine the Molar Mass of Aluminium (Al):
[tex]\[ \text{Molar Mass of Al} = 26.98 \, \text{g/mol} \][/tex]
2. Calculate the Moles of Aluminium (Al) Used:
Given mass of aluminium is 12.8 g.
[tex]\[ \text{Moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} \][/tex]
[tex]\[ \text{Moles of Al} = \frac{12.8 \, \text{g}}{26.98 \, \text{g/mol}} \][/tex]
[tex]\[ \text{Moles of Al} \approx 0.474 \, \text{mol} \][/tex]
3. Determine the Stoichiometric Ratio Between Aluminium and HCl:
From the balanced chemical equation, 2 moles of Al react with 6 moles of HCl. Therefore, 1 mole of Al reacts with 3 moles of HCl.
[tex]\[ \text{Moles of HCl} = \text{Moles of Al} \times 3 \][/tex]
[tex]\[ \text{Moles of HCl} = 0.474 \, \text{mol} \times 3 \][/tex]
[tex]\[ \text{Moles of HCl} \approx 1.423 \, \text{mol} \][/tex]
4. Determine the Molar Mass of Hydrochloric Acid (HCl):
[tex]\[ \text{Molar Mass of HCl} = 36.46 \, \text{g/mol} \][/tex]
5. Calculate the Mass of HCl Required:
[tex]\[ \text{Mass of HCl} = \text{moles of HCl} \times \text{molar mass of HCl} \][/tex]
[tex]\[ \text{Mass of HCl} = 1.423 \, \text{mol} \times 36.46 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of HCl} \approx 51.9 \, \text{g} \][/tex]
### Conclusion:
The mass of HCl needed to react completely with 12.8 g of aluminium is approximately 51.9 g. Therefore, the correct answer is:
[tex]\[ \boxed{51.9 \, \text{g}} \][/tex]
