A 55.0-mg sample of Al(OH)3 is reacted with 0.200M HCl. How many milliters of the acid are needed to neutralize the Al(OH)3.
The equation goes like this. (Un-labeled)
0.055g x (1 mol)/(78g) x (3 mol)/(1 mol) x (1L)/(0.200M) = 0.01057
You then convert this to mL
0.01057 * 1000 = (10.6 mL HCL) <-- rounded to sf
Hope this helps! (10 years later) lol
To determine the number of milliliters of 0.200M HCl required to neutralize the Al(OH)3, we need to first calculate the number of moles of Al(OH)3 and then use stoichiometry to find the volume of HCl.
1. Calculate moles of Al(OH)3:
- 1 mole of Al(OH)3 = molar mass of Al(OH)3 = (27.0 g/mol) + 3*(16.0 g/mol + 1.0 g/mol) = 78.0 g/mol.
- Number of moles of Al(OH)3 = mass / molar mass.
- Number of moles of Al(OH)3 = 55.0 mg / 78.0 g/mol = 0.705 mol.
2. Apply stoichiometry:
- Balanced equation: Al(OH)3 + 3HCl → AlCl3 + 3H2O.
- From the equation, we see that 1 mole of Al(OH)3 reacts with 3 moles of HCl.
- Therefore, the moles of HCl required = number of moles of Al(OH)3 * (3 moles HCl / 1 mole Al(OH)3).
- Moles of HCl required = 0.705 mol * (3 mol HCl / 1 mol Al(OH)3) = 2.115 mol.
3. Convert moles of HCl to volume in milliliters:
- Molarity (M) = moles (n) / volume (V in liters).
- Rearranging the equation, volume (V) = moles (n) / Molarity (M).
- Volume of HCl = moles of HCl required / Molarity of HCl.
- Volume of HCl = 2.115 mol / 0.200 mol/L = 10.575 L = 10575 mL.
Therefore, 10575 milliliters of 0.200M HCl are required to neutralize the 55.0-mg sample of Al(OH)3.
To find the volume of HCl required to neutralize the Al(OH)3, we can first calculate the number of moles of Al(OH)3 in the given sample, and then use stoichiometry to determine the number of moles of HCl required.
Here's the step-by-step solution:
1. Calculate the number of moles of Al(OH)3:
- Given mass of Al(OH)3 = 55.0 mg
- Convert the mass to grams:
55.0 mg = 55.0 × 10^-3 g (since 1 mg = 10^-3 g)
- Divide the mass by the molar mass of Al(OH)3:
Molar mass of Al(OH)3 = 3(Al) + 3(O) + 3(H) = 26.98 + 3(16.00) + 3(1.01) ≈ 78.00 g/mol
Moles of Al(OH)3 = (55.0 × 10^-3 g) / (78.00 g/mol)
2. Use the balanced chemical equation to determine the stoichiometry between Al(OH)3 and HCl:
2 Al(OH)3 + 6 HCl → 2 AlCl3 + 6 H2O
According to the equation, 2 moles of Al(OH)3 react with 6 moles of HCl.
3. Determine the number of moles of HCl needed:
Moles of HCl = (moles of Al(OH)3) × (6 mol HCl / 2 mol Al(OH)3)
4. Finally, calculate the volume of HCl in milliliters:
Given HCl concentration = 0.200 M (0.200 mol/L)
Volume of HCl = (Moles of HCl) / (HCl concentration)
Plug in the calculated numbers to get the final answer.
Emily, this is worked almost the same way as the NaOH/H2SO4 problem I worked for you earlier. What is it you don't understand about this?
All of them use this standard format.
1. Convert from what you have to mols.
2. Convert mols of what you have to mols of what you want.
3. Convert mols of what you want to whatever units you want; i.e., grams, liters, etc.