Redox and Equivalent ConceptHard

Question

When 0.91 g of a mixture of Na2SO4 and (NH4)2SO4 was boiled with 80 ml of 0.1 N-NaOH until no more NH3 is evolved, the excess of NaOH required is 11.6 ml of 0.1 N-HCl. How many grams of Na2SO4 is present in the mixture?

Options

A.0.594 g
B.0.459 g
C.0.549 g
D.0.945 g

Solution

$n_{eq}NaOH = n_{eq}\left( NH_{4} \right)_{2}SO_{4} + n_{eq}HCl$

Or $\frac{80 \times 0.1}{1000} = \frac{w}{132} \times 2 + \frac{11.6 \times 0.1}{1000} \Rightarrow w = 0.451\text{ g}$

∴ Mass of Na2SO4 = 0.91 – 0.451 = 0.459 g

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