Find the volume of 0.120 M hydrochloric acid necessary to react completely with 1.51 g Al(OH)^3
3HCl + Al(OH)3 ==> 3H2O + AlCl3
moles Al(OH)3 = grams/molar mass
Using the coefficients in the balanced equation, convert moles Al(OH)3 to moles HCl.
Then M HCl = moles HCl/L HCl. You know M and moles HCl, solve for L.
.1613
0.4839L
To find the volume of hydrochloric acid necessary to react completely with Al(OH)3, we first need to determine the amount of moles of Al(OH)3 using its molar mass and mass given.
Step 1: Calculate the molar mass of Al(OH)3.
Aluminum (Al) atomic mass = 26.98 g/mol
Oxygen (O) atomic mass = 16.00 g/mol
Hydrogen (H) atomic mass = 1.01 g/mol
Multiply the atomic masses by their respective subscripts:
Al: 26.98 g/mol * 1 = 26.98 g/mol
O: 16.00 g/mol * 3 = 48.00 g/mol
H: 1.01 g/mol * 3 = 3.03 g/mol
Adding these values together, we get:
26.98 g/mol + 48.00 g/mol + 3.03 g/mol = 77.01 g/mol
Step 2: Calculate the moles of Al(OH)3.
Moles = Mass / Molar mass
Moles = 1.51 g / 77.01 g/mol
Moles ≈ 0.0196 mol
Step 3: The balanced chemical equation for the reaction between Al(OH)3 and HCl is:
2 Al(OH)3 + 6 HCl → 2 AlCl3 + 6 H2O
From the equation, we can see that 2 moles of Al(OH)3 react with 6 moles of HCl.
Step 4: Determine the moles of HCl needed to react with Al(OH)3.
From the equation, the ratio is 2 moles Al(OH)3 to 6 moles HCl.
Moles of HCl = (Moles of Al(OH)3) × (6 moles HCl / 2 moles Al(OH)3)
Moles of HCl = 0.0196 mol × (6 / 2)
Moles of HCl = 0.0588 mol
Step 5: Calculate the volume of hydrochloric acid.
To calculate the volume, we need the concentration of hydrochloric acid.
The concentration of hydrochloric acid is given as 0.120 M, which means there are 0.120 moles of HCl in 1 liter of solution.
Volume (L) = Moles / Concentration
Volume (L) = 0.0588 mol / 0.120 mol/L
Volume (L) = 0.49 L
Therefore, 0.120 M hydrochloric acid necessary to react completely with 1.51 g Al(OH)3 is approximately 0.49 liters.