NaOH(s)+ H2SO4(aq)=Na2SO4(aq) + H2O(l)
Consider the unbalanced equation above.A 0.900 g sample of impure NaOH was dissolved in water and required 37.0 mL of 0.145 M H2SO4 solution to react with the NaOH in the sample. What was the mass percent of NaOH in the sample?
Balance the equation.
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
moles H2SO4 = M x L = 0.145 x 0.037 = 0.005365
Now look at the equation. There are two moles NaOH for each mole H2SO4; therefore, moles NaOH titrated = 2 x 0.005365 = 0.01073
grams NaOH = moles x molar mass = 0.01073 x 40 (approximately--you use the more exact number) = 0.4292
%NaOH = (grams NaOH/mass sample)*100 = ??, then round to 3 significant figures.
Check my math.
How do you get the mass of the entire sample?
To find the mass percent of NaOH in the sample, we need to determine the amount of NaOH reacted with the H2SO4 solution.
Step 1: Calculate the number of moles of H2SO4 used.
First, we calculate the number of moles of H2SO4 by using the concentration and volume provided:
n(H2SO4) = C(H2SO4) x V(H2SO4)
n(H2SO4) = 0.145 M x 0.037 L = 0.005365 moles
Step 2: Write the balanced chemical equation for the reaction.
The balanced equation for the reaction between NaOH and H2SO4 is:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
Step 3: Determine the stoichiometry of the reaction.
The balanced equation shows that 2 moles of NaOH react with 1 mole of H2SO4. Therefore, the moles of NaOH reacted will be half the moles of H2SO4 used:
n(NaOH) = 0.005365 moles / 2 = 0.002682 moles
Step 4: Calculate the molar mass of NaOH.
NaOH has a molar mass of:
M(NaOH) = 22.99 g/mol (Na) + 16.00 g/mol (O) + 1.01 g/mol (H) = 39.99 g/mol
Step 5: Determine the mass of NaOH reacted.
The mass of NaOH reacted can be calculated using the molar mass and the number of moles:
m(NaOH) = n(NaOH) x M(NaOH) = 0.002682 moles x 39.99 g/mol = 0.1074 g
Step 6: Calculate the mass percent of NaOH.
Finally, we can calculate the mass percent of NaOH in the sample using the mass of NaOH reacted and the initial mass of the sample:
mass percent = (mass of NaOH / mass of sample) x 100%
mass percent = (0.1074 g / 0.900 g) x 100% = 11.93%
Therefore, the mass percent of NaOH in the sample is approximately 11.93%.
To find the mass percent of NaOH in the sample, we need to determine the moles of NaOH and the total mass of the sample.
First, let's calculate the moles of H2SO4 used in the reaction:
Molarity of H2SO4 = 0.145 M
Volume of H2SO4 used = 37.0 mL = 0.0370 L
Moles of H2SO4 = Molarity x Volume
= 0.145 mol/L x 0.0370 L
= 0.005365 mol
According to the balanced equation, the stoichiometric ratio between H2SO4 and NaOH is 1:2. Therefore, the ratio between the moles of H2SO4 and NaOH is also 1:2.
Moles of NaOH = (2 x Moles of H2SO4)
= 2 x 0.005365 mol
= 0.01073 mol
Now let's calculate the molar mass of NaOH:
Molar mass of NaOH = (1 x atomic mass of Na) + (1 x atomic mass of O) + (1 x atomic mass of H)
= (1 x 22.99 g/mol) + (1 x 16.00 g/mol) + (1 x 1.01 g/mol)
= 22.99 g/mol + 16.00 g/mol + 1.01 g/mol
= 40.00 g/mol
Next, we can calculate the mass of NaOH in the sample:
Mass of NaOH = Moles of NaOH x Molar mass of NaOH
= 0.01073 mol x 40.00 g/mol
= 0.429 g
Finally, we can calculate the mass percent of NaOH in the sample:
Mass percent of NaOH = (Mass of NaOH / Mass of sample) x 100%
The mass of the sample is given as 0.900 g:
Mass percent of NaOH = (0.429 g / 0.900 g) x 100%
= 47.67%
Therefore, the mass percent of NaOH in the sample is approximately 47.67%.