To solve this problem, let's carefully go through the given reaction and the quantities involved step by step.
1. Identify the Molar Masses:
- The molar mass of [tex]\( Al_2(SO_4)_3 \)[/tex] is given as [tex]\( 342.15 \, \text{g/mol} \)[/tex].
- The molar mass of [tex]\( HCl \)[/tex] is given as [tex]\( 36.46 \, \text{g/mol} \)[/tex].
2. Convert the Volume of [tex]\( HCl \)[/tex] to Liters:
- The volume of [tex]\( HCl \)[/tex] solution is [tex]\( 500 \, \text{mL} \)[/tex], which is equivalent to [tex]\( 0.500 \, \text{L} \)[/tex].
3. Calculate the Moles of [tex]\( Al_2(SO_4)_3 \)[/tex]:
- Given mass of [tex]\( Al_2(SO_4)_3 \)[/tex] is [tex]\( 76.3 \, \text{g} \)[/tex].
- Moles of [tex]\( Al_2(SO_4)_3 \)[/tex] = [tex]\(\frac{\text{mass}}{\text{molar mass}} = \frac{76.3 \, \text{g}}{342.15 \, \text{g/mol}} = 0.223 \, \text{mol}\)[/tex].
4. Calculate the Moles of [tex]\( HCl \)[/tex]:
- Molarity of [tex]\( HCl \)[/tex] is [tex]\( 3.16 \, \text{M} \)[/tex].
- Volume of [tex]\( HCl \)[/tex] is [tex]\( 0.500 \, \text{L} \)[/tex].
- Moles of [tex]\( HCl \)[/tex] = [tex]\(\text{molarity} \times \text{volume} = 3.16 \, \text{mol/L} \times 0.500 \, \text{L} = 1.58 \, \text{mol}\)[/tex].
5. Relate the Reactants via Stoichiometry:
- The balanced reaction shows that [tex]\( 1 \)[/tex] mole of [tex]\( Al_2(SO_4)_3 \)[/tex] reacts with [tex]\( 6 \)[/tex] moles of [tex]\( HCl \)[/tex] to produce [tex]\( 3 \)[/tex] moles of [tex]\( H_2SO_4 \)[/tex].
- Therefore, moles of [tex]\( HCl \)[/tex] needed = [tex]\( \text{moles of } Al_2(SO_4)_3 \times 6 = 0.223 \times 6 = 1.34 \, \text{mol} \)[/tex].
6. Calculate the Moles of [tex]\( H_2SO_4 \)[/tex] Produced:
- From the stoichiometric relation, [tex]\( 1 \)[/tex] mole of [tex]\( Al_2(SO_4)_3 \)[/tex] produces [tex]\( 3 \)[/tex] moles of [tex]\( H_2SO_4 \)[/tex].
- Moles of [tex]\( H_2SO_4 \)[/tex] produced = [tex]\( \text{moles of } Al_2(SO_4)_3 \times 3 = 0.223 \times 3 = 0.669 \, \text{mol} \)[/tex].
7. Calculate the Final Molarity of [tex]\( H_2SO_4 \)[/tex]:
- Assume the volume of the solution remains unchanged at [tex]\( 0.500 \, \text{L} \)[/tex].
- Final molarity of [tex]\( H_2SO_4 \)[/tex] = [tex]\(\frac{\text{moles of } H_2SO_4}{\text{volume of solution}} = \frac{0.669 \, \text{mol}}{0.500 \, \text{L}} = 1.338 \, \text{M}\)[/tex].
Therefore, the final molarity of [tex]\( H_2SO_4 \)[/tex] is [tex]\( 1.338 \, \text{M} \)[/tex].
