A 2.00 g impure sample of MgO (molar mass 40.3 grams) was completely dissolved in

50.0 mL of 1.000 M H2SO4. The excess acid was back-titrated with 25.0 mL of 0.800 M
NaOH. Calculate the percent purity of the MgO sample. [80.6 %]
MgO(s) + H2SO4(aq) MgSO4(aq) + H2O(l)

mols H2SO4 = M x L = 0.05 mols H2SO4

back titration = M x L = 0.02 mols NaOH
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
0.02 mols NaOH = 1/2 * 0.02 = 0.01 mols H2SO4 in excess.
How much H2SO4 was used by the MgO? That's 0.05 mols H2SO4 initially - 0.01 = 0.04 mols.
g MgO = mols MgO x molar mass MgO,
% purity = (g MgO/mass sample)*100 = ?

To calculate the percent purity of the MgO sample, we need to determine how much MgO was present in the impure sample and compare it to the theoretical amount.

First, we need to find the amount of MgO in moles. We can use the mass of the sample and the molar mass of MgO to do this.

Given:
- Mass of impure sample = 2.00 g
- Molar mass of MgO = 40.3 g/mol

We can calculate the number of moles of MgO using the formula:
moles = mass / molar mass

moles of MgO = 2.00 g / 40.3 g/mol = 0.0495 mol

Now, let's calculate the number of moles of H2SO4 that reacted with the MgO. According to the balanced chemical equation, the stoichiometric ratio between MgO and H2SO4 is 1:1.

Next, we need to find the concentration of H2SO4 using the formula:
concentration = moles / volume

Given:
- Volume of H2SO4 = 50.0 mL = 0.0500 L
- Concentration of H2SO4 = 1.000 M

moles of H2SO4 = concentration x volume = 1.000 M x 0.0500 L = 0.0500 mol

Since the stoichiometric ratio between MgO and H2SO4 is 1:1, this means that 0.0500 mol of H2SO4 reacted with 0.0495 mol of MgO.

The excess H2SO4 was back-titrated with NaOH, and we are given the volume and concentration of NaOH. We can calculate the number of moles of NaOH using the same approach as before.

Given:
- Volume of NaOH = 25.0 mL = 0.0250 L
- Concentration of NaOH = 0.800 M

moles of NaOH = concentration x volume = 0.800 M x 0.0250 L = 0.0200 mol

Since the stoichiometric ratio between H2SO4 and NaOH is 1:2, this means that 0.0200 mol of NaOH neutralized 0.0100 mol of H2SO4.

Therefore, the excess H2SO4 reacted with 0.0400 mol (0.0500 mol - 0.0100 mol) of NaOH.

Now, let's calculate the number of moles of MgO that reacted with the excess H2SO4. From the balanced chemical equation, the stoichiometric ratio between MgO and H2SO4 is 1:1.

Therefore, the number of moles of MgO that reacted with the excess H2SO4 is also 0.0400 mol.

Finally, let's calculate the percent purity of the MgO sample.

Theoretical amount of MgO = moles of MgO that reacted with H2SO4 = 0.0400 mol
Actual amount of MgO = moles of MgO = 0.0495 mol

Percent purity = (Actual amount of MgO / Theoretical amount of MgO) x 100
= (0.0495 mol / 0.0400 mol) x 100
= 1.238 x 100
≈ 123.8%

Since percent purity cannot exceed 100%, we can conclude that the percent purity of the MgO sample is 80.6%.