Determine how many milliliters of 0.448 M HClO4 will be required to neutralize 38.30 g Ca(OH)2 according to the reaction: 2HCl+Ca(ClO4)2+2H20
You need to start from a balanced equation
HClO4 + Ca(OH)2 -> Ca(ClO4)2 + H20
which you need to balance
2HClO4 + Ca(OH)2 -> Ca(ClO4)2 + 2H20
then work out how many moles 38.30 g Ca(OH)2 represents from
number of moles = 38.30 g/molar mass
from the equation 2 mole of HClO4 needed for each mole of Ca(OH)2
hence calcutae the number of moles of HClO4 needed.
ml of HClO4 needed is then
number of moles x 1000/(0.448 mole L^-1)
To determine how many milliliters of 0.448 M HClO4 are required to neutralize 38.30 g Ca(OH)2, we need to use stoichiometry and the balanced chemical equation.
First, let's find the molar mass of Ca(OH)2. Calcium (Ca) has a molar mass of 40.08 g/mol, while each hydroxide (OH) group has a molar mass of 17.01 g/mol (1 oxygen atom: 15.999 g/mol + 1 hydrogen atom: 1.01 g/mol). So, the molar mass of Ca(OH)2 is:
Ca: 1 × 40.08 g/mol = 40.08 g/mol
(OH)2: 2 × (17.01 g/mol) = 34.02 g/mol
Total molar mass of Ca(OH)2 = 40.08 g/mol + 34.02 g/mol = 74.10 g/mol
Now, let's use stoichiometry to calculate the number of moles of Ca(OH)2.
Moles of Ca(OH)2 = mass / molar mass
Moles of Ca(OH)2 = 38.30 g / 74.10 g/mol
Moles of Ca(OH)2 ≈ 0.5177 mol
Looking at the balanced chemical equation:
2 HCl + Ca(ClO4)2 + 2 H2O → 2 Ca(ClO4) + 4 H2O
We see that the stoichiometric ratio between HClO4 and Ca(OH)2 is 2:1. This means that 2 moles of HClO4 will react with 1 mole of Ca(OH)2.
Since we have 0.5177 mol of Ca(OH)2, we will need:
Moles of HClO4 = (0.5177 mol Ca(OH)2) × (2 mol HClO4 / 1 mol Ca(OH)2)
Moles of HClO4 ≈ 1.0354 mol
Now, we can calculate the volume (in milliliters) of 0.448 M HClO4 solution needed. The molar concentration (M) of a solution represents the number of moles of solute per liter of solution. Since we have moles of HClO4, we can use the molar concentration to calculate the volume.
Volume of HClO4 (in liters) = Moles of HClO4 / Molarity of HClO4
Volume of HClO4 (in liters) = 1.0354 mol / 0.448 mol/L
Volume of HClO4 (in liters) ≈ 2.31 L
Finally, we can convert the volume from liters to milliliters:
Volume of HClO4 (in milliliters) = Volume of HClO4 (in liters) × 1000
Volume of HClO4 (in milliliters) ≈ 2.31 L × 1000
Volume of HClO4 (in milliliters) ≈ 2310 mL
Therefore, approximately 2310 milliliters of 0.448 M HClO4 will be required to neutralize 38.30 g Ca(OH)2.