Final answer:
To find the volume of 1.24 M H2SO4 required to react with 1.26 g of Al, we can use stoichiometry. The volume of 1.24 M H2SO4 required is 56.5 mL.
Step-by-step explanation:
To find the volume of 1.24 M H2SO4 required to react with 1.26 g of Al, we need to use stoichiometry. First, calculate the number of moles of Al using its molar mass. Then, use the balanced equation to determine the mole ratio between Al and H2SO4. Finally, convert the moles of H2SO4 to volume using its molarity.
The molar mass of Al is 26.98 g/mol. The number of moles of Al is therefore 1.26 g / 26.98 g/mol = 0.0467 mol. From the balanced equation, we can see that the mole ratio between Al and H2SO4 is 2:3. So, the number of moles of H2SO4 is (0.0467 mol Al) x (3 mol H2SO4 / 2 mol Al) = 0.07 mol H2SO4.
Now we can use the molarity of H2SO4 to find its volume. Molarity = moles / volume. Rearranging the formula, volume = moles / molarity. The volume of 1.24 M H2SO4 required is therefore 0.07 mol / 1.24 M = 56.5 mL.