Final answer:
To find the volume of hydrochloric acid necessary to react completely with aluminum hydroxide, we use stoichiometry. Given the concentration of HCl and the mass of Al(OH)3, we can calculate the volume of HCl solution needed. The volume is 0.05888 L or 58.88 mL.
Step-by-step explanation:
To find the volume of hydrochloric acid necessary to react completely with aluminum hydroxide, we need to use stoichiometry. First, we need to write the balanced chemical equation for the reaction:
Al(OH)3 + 3HCl → AlCl3 + 3H2O
From the equation, we can see that 1 mole of Al(OH)3 reacts with 3 moles of HCl. Given that the concentration of HCl is 0.297 mol/L, we can calculate the moles of HCl required:
Moles of HCl = (0.297 mol/L) x (volume in L)
To find the volume of HCl solution needed to react completely with 1.52 g of Al(OH)3, we need to convert the mass of Al(OH)3 to moles:
Moles of Al(OH)3 = 1.52 g / (molar mass of Al(OH)3)
Since the molar mass of Al(OH)3 is 78.00 g/mol, we can calculate the moles:
Moles of Al(OH)3 = 1.52 g / 78.00 g/mol
Now, we can use the mole ratio from the balanced equation to find the volume of HCl solution:
volume of HCl solution = (moles of Al(OH)3) x (volume of HCl solution per mole of Al(OH)3)
Substituting the values, we get:
volume of HCl solution = (1.52 g / 78.00 g/mol) x (3 mol HCl / 1 mol Al(OH)3) x (1 L / 0.297 mol)
Simplifying the equation, we find:
volume of HCl solution = 0.05888 L
Therefore, the volume of 0.297 mol/L hydrochloric acid necessary to react completely with 1.52 g of Al(OH)3 is 0.05888 L or 58.88 mL.