How many milliliters of 4.80 M HCl are required to react with 110 mL of 1.48 M Al(OH)3?
Al(OH)3(s) + 3HCl(aq)to AlCl3(aq) + 3H2O(aq)
mols Al(OH)3 = M x L = ?
Convert mols Al(OH)3 to mols HCl.
Then M HCl = mols HCl/L HCl. You know mols and M, solve for L and convert to mL.
To find the number of milliliters of a solution required to react with another solution, you need to use the stoichiometry of the balanced chemical equation. In this case, the balanced chemical equation is:
Al(OH)3(s) + 3HCl(aq) → AlCl3(aq) + 3H2O(aq)
From the balanced equation, you can see that the stoichiometric ratio between Al(OH)3 and HCl is 1:3. This means that for every 1 mole of Al(OH)3, you need 3 moles of HCl to react completely.
First, calculate the number of moles of Al(OH)3 in 110 mL of a 1.48 M solution:
Moles of Al(OH)3 = Volume (in liters) × Concentration
= 110 mL × (1 L / 1000 mL) × 1.48 M
= 0.16328 moles
Since the stoichiometric ratio is 1:3 for Al(OH)3 and HCl, you need 3 times the number of moles of HCl:
Moles of HCl = 3 × Moles of Al(OH)3
= 3 × 0.16328 moles
= 0.48984 moles
Now, you need to calculate the volume of the 4.80 M HCl solution needed to contain 0.48984 moles of HCl:
Volume (in liters) = Moles of HCl / Concentration
= 0.48984 moles / 4.80 M
= 0.10197 liters
Finally, convert the volume from liters to milliliters:
Volume (in milliliters) = Volume (in liters) × (1000 mL / 1 L)
= 0.10197 liters × 1000 mL
= 101.97 mL
Therefore, approximately 101.97 milliliters of 4.80 M HCl are required to react with 110 mL of 1.48 M Al(OH)3.