Final answer:
To react with 69mL of a 2.5M aluminum hydroxide solution, 0.96L of a 4.85M hydrochloric acid solution is required.
Step-by-step explanation:
To determine the volume of a 4.85M hydrochloric acid (HCl) solution required to react with 69mL of a 2.5M aluminum hydroxide (Al(OH)3) solution, we can use the stoichiometry of the balanced chemical equation.
The balanced equation for the reaction is:
2Al(OH)3 + 6HCl → Al2Cl6 + 6H2O
From the balanced equation, we can see that there is a 2:6 ratio between aluminum hydroxide and hydrochloric acid. This means that for every 2.5 moles of aluminum hydroxide, we need 6 times that moles of hydrochloric acid.
To calculate the moles of hydrochloric acid required, we first convert the given volume of aluminum hydroxide solution to moles:
69mL x (1L/1000mL) x (2.5 mol/L) = 0.1725 moles of Al(OH)3
Since the mole ratio between Al(OH)3 and HCl is 2:6, we can calculate the moles of HCl required:
0.1725 moles of Al(OH)3 x (6 moles of HCl/2 moles of Al(OH)3) = 0.5175 moles of HCl
Finally, we can calculate the volume of 4.85M HCl solution required:
moles of HCl = (volume of HCl solution) x (4.85 mol/L)
0.5175 moles of HCl = (volume of HCl solution) x (4.85 mol/L)
volume of HCl solution = 0.5175 moles of HCl / 4.85 mol/L = 0.1068 L = 106.8 mL
Therefore, the correct answer is option A. 0.96 L.