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Limiting Reagent or Reactant

11.2 g carbon...

Question

$11.2 g$ carbon reacts completely with $19.63 l i t r e$ $O_{2}$ at $N T P$ . The cooled gases are passed through $2 l i t r e$ of $2.5 N N a O H$ solution. Calculate concentration of remaining $N a O H$ and $N a_{2} C O_{3}$ in solution. ( $C O$ does not react with $N a O H$ under these conditions.)

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Solution

Let $x$ moles of carbon be converted into $C O$ and $y$ moles of carbon be converted into $C O_{2}$ .
$C x + \frac{1}{2} O_{2} x / 2 \to C O$
$C y + O_{2} y \to C O_{2}$
Total volume of oxygen used $= \frac{x}{2} \times 22.4 + y \times 22.4$
$= 19.63$
$11.2 x + 22.4 y = 19.63 . . . . (i)$
$x + y = \frac{11.2}{12}$ ; i.e., $12 x + 12 y = 11.2 . . . . (i i)$
Solving eqs. (i) and (ii), we get
$x = 0.11, y = 0.82$
Number of moles of $C O_{2}$ formed $= 0.82$
Number of milliequivalents of $N a O H$ solution through which $O_{2}$ is massed $= N \times V = 2.5 \times 2000 = 5000$
Number of milliequivalents of $C O_{2}$ passed $= 0.82 \times 2 \times 1000$
$= 1640$
$2 N a O H + C O_{2} \to N a_{2} C O_{3} + H_{2} O$
Number of milliequivalents of $N a_{2} C O_{3} = 1640$
$N_{N a_{2} C O_{3}} = \frac{1640}{2000} = 0.82$
Number of milliequivalent of remaining $N a O H$
$= 5000 - 1640 = 3360$
Normality of remaining $N a O H \frac{3360}{2000} = 1.68$ .

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11.2 g carbon reacts completely with 19.63 litre O2 at NTP. The cooled gases are passed through 2 litre of 2.5 N NaOH solution. Calculate concentration of remaining NaOH and Na2CO3 in solution. (CO does not react with NaOH under these conditions.)