# A 0.205 g sample of impure NaOH requires 17.5 mL of 0.2180 M HCl for neutralization. What is the percent of NaOH in the sample, by weight?

## moles acid=.2180*.0175

mass NaOH=formulamassNaOH*molesacid
mass NaOH=40*moles acid above

percent= massNaOH/.205

## To find the percent of NaOH in the sample by weight, we need to determine the amount of NaOH in the sample and then calculate its percentage.

Step 1: Calculate the number of moles of HCl used.
To find the number of moles of HCl, we can use the formula:
moles = concentration (M) × volume (L)

moles of HCl = 0.2180 M × 0.0175 L = 0.003815 mol

Step 2: Write the balanced chemical equation for the reaction between NaOH and HCl.
NaOH + HCl → NaCl + H2O

This equation tells us that 1 mole of HCl reacts with 1 mole of NaOH.

Step 3: Use the stoichiometry of the balanced equation to find the number of moles of NaOH.
Since the stoichiometry of the reaction is 1:1, the number of moles of NaOH is also 0.003815 mol.

Step 4: Calculate the molar mass of NaOH.
The molar mass of NaOH is:
Na = 22.99 g/mol
O = 16.00 g/mol
H = 1.01 g/mol

Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.00 g/mol

Step 5: Calculate the mass of NaOH in the sample.
The mass of NaOH in the sample is:
mass = moles × molar mass

mass of NaOH = 0.003815 mol × 39.00 g/mol = 0.14839 g

Step 6: Calculate the percentage of NaOH in the sample.
The percentage of NaOH in the sample is:
percentage = (mass of NaOH / mass of sample) × 100

percentage of NaOH = (0.14839 g / 0.205 g) × 100 = 72.34%

Therefore, the percent of NaOH in the sample, by weight, is approximately 72.34%.