NaOH(s)+ H2SO4(aq)=Na2SO4(aq) + H2O(l)

Question

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?

Answer 1

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.

Answer 2

How do you get the mass of the entire sample?