Determine how many milliliters of 0.255 M HClO4 will be required to neutralize 48.56 g Ca(OH)2 according to the reaction:

2HClO4(aq)+Ca(OH)2(s)=CA(ClO4)2(aq)+2H2O(l)

Here is the corrected equation.

2HClO4(aq)+Ca(OH)2(s)=Ca(ClO4)2(aq)+2H2O(l)
mols Ca(OH)2 = grams/molar mass
Using the coefficients in the balanced equation, convert mols Ca(OH)2 to mols HClO4.
Then M HClO4 = mols HClO4/L HClO4. You know Mols and M, solver for L HClO4 and convert to mL.

I love chemistry sm ๐Ÿงช๐Ÿ‘จโ€๐Ÿ”ฌ๐Ÿฅผ

To determine the volume of 0.255 M HClO4 required to neutralize 48.56 g of Ca(OH)2, we need to use the stoichiometry of the balanced equation.

First, let's calculate the number of moles of Ca(OH)2 in 48.56 g:

Step 1: Calculate the molar mass of Ca(OH)2.
Ca: 1 atom x 40.08 g/mol = 40.08 g/mol
O: 2 atoms x 16.00 g/mol = 32.00 g/mol
H: 2 atoms x 1.01 g/mol = 2.02 g/mol

Total molar mass of Ca(OH)2 = 40.08 g/mol + 32.00 g/mol + 2.02 g/mol = 74.10 g/mol

Step 2: Calculate the number of moles of Ca(OH)2.
48.56 g / 74.10 g/mol = 0.6556 mol

Now, let's use the stoichiometry of the balanced equation to find the number of moles of HClO4 required.

From the balanced equation:
2 moles of HClO4 react with 1 mole of Ca(OH)2.

Step 3: Calculate the number of moles of HClO4 required.
0.6556 mol of Ca(OH)2 x (2 mol of HClO4 / 1 mol of Ca(OH)2) = 1.3112 mol of HClO4

Finally, let's convert the moles of HClO4 to milliliters (mL) using the given concentration of HClO4.

Step 4: Calculate the volume of 0.255 M HClO4 required.
Volume (L) = moles / concentration (M)
Volume (L) = 1.3112 mol / 0.255 M = 5.137 L

Now, convert liters to milliliters:
Volume (mL) = 5.137 L x 1000 mL/L = 5137 mL

Therefore, approximately 5137 mL of 0.255 M HClO4 will be required to neutralize 48.56 g of Ca(OH)2.

To determine how many milliliters of 0.255 M HClO4 are required to neutralize 48.56 g of Ca(OH)2, we need to use stoichiometry and the given balanced equation.

First, we need to calculate the number of moles of Ca(OH)2 using its molar mass. The molar mass of Ca(OH)2 is calculated by adding up the atomic masses of calcium (Ca), two hydrogens (H), and two oxygens (O):

Molar mass of Ca(OH)2 = atomic mass of Ca + (2 * atomic mass of H) + (2 * atomic mass of O)
= 40.08 g/mol + (2 * 1.01 g/mol) + (2 * 16.00 g/mol)
= 40.08 g/mol + 2.02 g/mol + 32.00 g/mol
= 74.10 g/mol

Next, we can calculate the number of moles of Ca(OH)2 by dividing the mass given (48.56 g) by its molar mass:

Number of moles of Ca(OH)2 = mass of Ca(OH)2 / molar mass of Ca(OH)2
= 48.56 g / 74.10 g/mol
โ‰ˆ 0.655 moles

According to the balanced equation, the ratio between HClO4 and Ca(OH)2 is 2:1. Therefore, the number of moles of HClO4 required to neutralize the given amount of Ca(OH)2 is:

Number of moles of HClO4 = (0.655 moles of Ca(OH)2) * (2 moles of HClO4 / 1 mole of Ca(OH)2)
= 1.31 moles

Lastly, we can use the molarity (0.255 M) of HClO4 to calculate the volume of HClO4 required:

Volume of HClO4 = number of moles of HClO4 / molarity of HClO4
= 1.31 moles / 0.255 mol/L
โ‰ˆ 5.14 L (since 1 L = 1000 mL)

Therefore, approximately 5.14 mL of 0.255 M HClO4 will be required to neutralize 48.56 g of Ca(OH)2.