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Stoichiometric Calculations

The gases pro...

Question

The gases produced when $18$ g carbon reacts with $5$ L of oxygen at $18$ and $5$ atm pressure are treated with $0.5$ L of $2 M N a O H$ . The concentration of sodium carbonate produced by the reaction of $C O_{2}$ with $N a O H$ is_________. (as nearest integer)
( $C O$ does not react under these conditions).

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Solution

$C + \frac{1}{2} O_{2} \to C O . . . (i)$
$C + O_{2} \to C O_{2} . . (i i)$
let a mole of C reacts according to Eq. (i) and b mole of c reacts according to Eq. (ii).
Then, $a + b = \frac{18}{12} = 1.5 . . . (i i i)$
Also, Mole of $O_{2}$ used = $\frac{a}{2} + b = \frac{5 \times 5}{0.0821 \times 291} (n = \frac{P V}{R T})$
or, $\frac{a}{2} + b = 1.046 . . . (i v)$
By Eqs. (iii) and (iv) $a = 0.908, b = 0.592$
$∴ C O_{2} f o r m e d = 0.592 m o l$
Now , $2 N a O H + C O_{2} \to {N a}_{2} C O_{3} . . . (v)$
$2 N a O H + C O_{2} \to {N a}_{2} C O_{3} + H_{2} O$
1.0 0.592 0 0
0 0.092 0.5 _
Thus, Mole of ${N a}_{2} C O_{3}$ formed $= 0.5$
Also,Mole of $C O_{2}$ left $= 0.092$
Now, ${N a}_{2} C O_{3} + C O_{2} + H_{2} O \to 2 N a H C O_{3}$
0.5 0.092 0
0.408 0 0.184
Mole of ${N a}_{2} C O_{3}$ left $= 0.408$
Mole of $N a H C O_{3}$ formed $= 0.148$
$M_{{N a}_{2} C O_{3}} = \frac{0.408}{0.5} = 0.816 M$
so nearest integer is $1$ .
$M_{N a H C O_{3}} = \frac{0.184}{0.5} = 0.368 M$

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