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
Thank you! That makes much more sense.
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%.