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Arrhenius Acid

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Question

Find the number of moles of $N a O H$ required to neutralize 448 g equimolar mixture of $N a_{2} C_{2} O_{4}$ and $H_{2} C_{2} O_{4}$ .

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Solution

Given, total mass of $N a_{2} C_{2} O_{4}$ and $H_{2} C_{2} O_{4}$ be 448 g.
Let, moles of each be x.
So, $x \times 134 + x \times 90 = 448 \Rightarrow x = 2$
Equivalents of $N a O H$ required
= Equivalent of $H_{2} C_{2} O_{4}$ (only for $H_{2} C_{2} O_{4})$
Moles of $N a O H$ $\times$ n-factor = mole of $H_{2} C_{2} O_{4}$ $\times$ n -factor
Moles of $N a O H = 2 \times 2 = 4$
(n-factor of $N a O H$ is 1 and n-factor of $H_{2} C_{2} O_{4}$ is 2)

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