A .055 gram sample of aluminum hydroxide is reacted with .200M hydrochloric acid. How many mL of the acid are needed to neutralize the aluminum hydroxide?
1. Write and balance the equation.
2. Convert 0.055 g Al(OH)3 to moles. moles = grams/molar mass.
3. Using the coefficients in the balanced equation, convert moles Al to moles HCl.
4. Molarity = moles HCl/L of HCl.
You know moles HCl and Molarity; calculate L of HCl.
To find out how many milliliters (mL) of hydrochloric acid are needed to neutralize the aluminum hydroxide, we first need to determine the number of moles of aluminum hydroxide present in the sample. Then, using the balanced chemical equation for the reaction between aluminum hydroxide and hydrochloric acid, we can determine the stoichiometry between the two compounds. Finally, we can use the concentration of hydrochloric acid to calculate the volume required.
1. Determine the number of moles of aluminum hydroxide:
To do this, we need to use the molar mass of aluminum hydroxide. The molecular formula of aluminum hydroxide is Al(OH)3, and the atomic masses are Al: 26.98 g/mol, O: 16.00 g/mol, and H: 1.01 g/mol.
Grams of aluminum hydroxide = 0.055 g
Molar mass of aluminum hydroxide = (26.98 g/mol) + 3(16.00 g/mol) + 3(1.01 g/mol) = 78.00 g/mol
Number of moles of aluminum hydroxide = (0.055 g) / (78.00 g/mol)
2. Determine the stoichiometry between aluminum hydroxide and hydrochloric acid:
The balanced chemical equation for the reaction between aluminum hydroxide (Al(OH)3) and hydrochloric acid (HCl) is:
Al(OH)3 + 3HCl -> AlCl3 + 3H2O
From the equation, we can see that 1 mole of aluminum hydroxide reacts with 3 moles of hydrochloric acid.
3. Use the concentration of hydrochloric acid to calculate the volume:
The concentration of hydrochloric acid is given as 0.200 M. This means that there are 0.200 moles of hydrochloric acid per liter of solution.
We can use the mole ratio from step 2 to calculate the volume of hydrochloric acid required to neutralize the aluminum hydroxide:
Volume of hydrochloric acid (in liters) = (Number of moles of aluminum hydroxide) * (3 moles HCl / 1 mole Al(OH)3)
Now, we can convert the volume from liters to milliliters:
Volume of hydrochloric acid (in mL) = (Volume of hydrochloric acid in liters) * (1000 mL / 1 L)
Simply plug in the values calculated in steps 1-3 to find the answer.