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Stoichiometric Calculations

NaOH can be p...

Question

$N a O H$ can be prepared by two methods each with two steps having $100 %$ extent.
$I$ : $2 N a + 2 H_{2} O ⟶ 2 N a O H + H_{2}$
$2 H_{2} + O_{2} ⟶ 2 H_{2} O$
$I I$ : $2 N a + \frac{1}{2} O_{2} ⟶ N a_{2} O$
$N a_{2} O + H_{2} O ⟶ 2 N a O H$
Which gives better yield?

I

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I I

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Both equal

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None is suitable

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Solution

The correct option is B Both equal
The reactions are as follows:
$I$ : $2 N a + 2 H_{2} O ⟶ 2 N a O H + H_{2}$
$2 H_{2} + O_{2} ⟶ 2 H_{2} O$
$I I$ : $2 N a + \frac{1}{2} O_{2} ⟶ N a_{2} O$
$N a_{2} O + H_{2} O ⟶ 2 N a O H$
According to equations, both require $2$ moles of $N a$ for $2$ moles of $N a O H$ .
So, they both give equal yield.

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