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Stoichiometry

1000 g of a m...

Question

1000 g of a mixture of $N a_{2} C O_{3}, N a_{2} S O_{4}$ and $N a O H$ for complete neutralisation requires 511 g of $H C l$ . The same mixture when reacted with excess of $B a C l_{2}$ solution, produces 466 g of white precipitate of $B a S O_{4}$ . Calculate the mass percentage of $N a O H$ in the mixture.
(Given Molar mass of $N a_{2} C O_{3} = 106 g, N a_{2} S O_{4} = 142 g, B a S O_{4} = 233 g, H C l = 36.5 g a n d N a O H = 40 g$ )

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Solution

Let, number of moles of $N a_{2} C O_{3} = a$
number of moles of $N a_{2} S O_{4} = b$
number of mole of $N a O H = c$
$a \times 106 + b \times 142 + c \times 40 = 1000$
Now, moles of $N a_{2} S O_{4} =$ moles of $B a S O_{4}$
$b = \frac{466}{233} = 2$
So, $106 a + 40 c = 1000 - 284 \Rightarrow 106 a + 40 c = 716$
For $H C l, a \times 2 + c \times 1 = \frac{511}{36.5} = 14 80 a + 40 c = 14 \times 40$
So, $26 a = 716 - 560 = 156 a = \frac{156}{26} = 6$
So, $c = 14 - 12 = 2$
So, weight of $N a O H = 2 \times 40 = 80 g$
% $N a O H = \frac{80}{1000} \times 100 = 8 %$

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