A 0.327 g sample of a mixture of NaOH(s) and NaCl(s) requires 31.5 mL of 0.150 M HCl(aq) to react with all the NaOH(s). What is the mass percentage of the NaOH(s) in the mixture?

HCl + NaOH ==> NaCl + H2O

mols HCl = M x L = ?
Look at the coefficients in the balanced equation. mols HCl = mols NaOH.
Then g NaOH = mols NaOH x molar mass NaOH = ? This is the theoretical yield.
%yield = (mass NaOH/mass sample)*100 = ?

To calculate the mass percentage of NaOH in the mixture, we need to determine the amount of NaOH that reacted with HCl.

First, we can find the moles of HCl used in the reaction using the formula:

moles of HCl = volume of HCl (in L) × molarity of HCl

Given that the volume of HCl used is 31.5 mL and the molarity is 0.150 M, we can convert the volume to liters:

Volume of HCl = 31.5 mL = 31.5 × 10^(-3) L

Next, we calculate the moles of HCl used:

moles of HCl = (31.5 × 10^(-3) L) × (0.150 mol/L) = 4.725 × 10^(-3) mol

According to the balanced chemical equation:

NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l)

The stoichiometric ratio between NaOH and HCl is 1:1. Therefore, the moles of NaOH present in the mixture are also 4.725 × 10^(-3) mol.

Now, we need to find the molar mass of NaOH, which is composed of sodium (Na) and hydroxide (OH). The molar mass of sodium (Na) is 22.99 g/mol, and the molar mass of oxygen (O) is 15.999 g/mol. Hydrogen (H) has a molar mass of 1.00784 g/mol.

Molar mass of NaOH = (22.99 g/mol + 15.999 g/mol + 1.00784 g/mol) = 39.99684 g/mol ≈ 40 g/mol

Finally, we can calculate the mass of NaOH in the mixture:

Mass of NaOH = moles of NaOH × molar mass of NaOH
Mass of NaOH = (4.725 × 10^(-3) mol) × (40 g/mol) ≈ 0.189 g

To find the mass percentage of NaOH in the mixture, divide the mass of NaOH by the total mass of the mixture and multiply by 100:

Mass % of NaOH = (0.189 g / 0.327 g) × 100 ≈ 57.82 %

Therefore, the mass percentage of NaOH in the mixture is approximately 57.82%.