How many milliliters of 0.291 M HCl would be required to titrate 6.83 g KOH?

mols KOH = grams/molar mass

mols HCl = mols KOH (look at the coefficients in the balanced equation).
M HCl = mol HCl/L HCl. Solve for L HCl and convert to mL.

To determine the number of milliliters of 0.291 M HCl required to titrate 6.83 g of KOH, we can use the concept of stoichiometry and the balanced equation of the reaction between HCl and KOH.

The balanced equation for the reaction between HCl and KOH is:

HCl + KOH → KCl + H2O

From the balanced equation, we can determine that the molar ratio between HCl and KOH is 1:1. This means that one mole of HCl reacts with one mole of KOH.

To determine the number of moles of KOH, we can use the molar mass of KOH, which is the sum of the atomic masses of potassium (K), oxygen (O), and hydrogen (H). The atomic masses are:

Potassium (K) = 39.10 g/mol
Oxygen (O) = 16.00 g/mol
Hydrogen (H) = 1.01 g/mol

Molar mass of KOH = 39.10 g/mol + 16.00 g/mol + 1.01 g/mol = 56.11 g/mol

Now, we can calculate the number of moles of KOH:

Number of moles of KOH = Mass of KOH / Molar mass of KOH
= 6.83 g / 56.11 g/mol
≈ 0.122 mol

Since the molar ratio between HCl and KOH is 1:1, the number of moles of HCl required to react with 0.122 moles of KOH is also 0.122 moles.

Now, we can determine the volume (in liters) of 0.291 M HCl required to contain 0.122 moles:

Volume (in liters) of HCl = Number of moles of HCl / Molarity of HCl
= 0.122 mol / 0.291 mol/L
≈ 0.419 L

Finally, we need to convert the volume from liters to milliliters. Since 1 liter is equal to 1000 milliliters:

Volume (in milliliters) of HCl = Volume (in liters) of HCl * 1000
≈ 0.419 L * 1000
≈ 419 mL

Therefore, approximately 419 milliliters of 0.291 M HCl would be required to titrate 6.83 g of KOH.