$18$ g $C$ and $40$ g $O_{2}$ on complete reaction gives gaseous mixture which is further treated with $2$ moles $N a O H$ solution. Calculate the moles of $N a_{2} C O_{3}$ formed (assuming no reaction of $N a O H$ occurs with $C O$ & complete reaction with $C O_{2}$ )

0.5

mol

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1

mol

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1.5

mol

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2

mol

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Solution

The correct option is B $1$ mol
$\begin{matrix} C + & \frac{1}{2} O_{2} \to & C O moles & a & \frac{a}{2} & a \end{matrix}$
$\begin{matrix} C + & O_{2} \to & C O_{2} moles & b & b & b \end{matrix}$

So according to the questions,
$a + b = \frac{18}{12} = 1.5 \cdot \cdot \cdot (i)$
$\frac{a}{2} + b = \frac{40}{32} = 1.25 \cdot \cdot \cdot (i i)$
Now, $(i) - (i i) \Rightarrow a = 0.5$
$∴ b = 1$
The corresponding chemical reaction of $C O_{2}$ and $N a O H$ is
$C O_{2} + 2 N a O H \to N a_{2} C O_{3} + H_{2} O$
$1 mol 2 mol$
So here $C O_{2}$ is the limiting reagent and thus $1 mol$ of $N a_{2} C O_{3}$ will be formed.

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