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 = ?
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)
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