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Stoichiometric Calculations

Chlorine diox...

Question

Chlorine dioxide $(C l O_{2})$ is used nowadays for water treatment rather than $C l_{2}$ . $C l O_{2}$ is obtained by passing $C l_{2} (g)$ into a concentrated solution of sodium chlorite $(N a C l O_{2})$ . The reaction gives $90 %$ yield. If $x$ moles of $C l O_{2}$ is produced in $3.78$ L of $2.0 M$ $N a C l O_{2} (a q)$ , then calculate the value of $10 x$ .

68

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60

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74

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None of the above

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Solution

The correct option is A $68$
$2 N a C l O_{2} 2 m o l + C l_{2} \to 2 N a C l + 2 C l O_{2} 2 m o l$
Moles of $C l O_{2}$ $=$ Moles of $N a C l O_{2}$ $=$ $M \times V_{L} = 2 \times 3.78$ $= 7.56$
Moles of $C l O_{2}$ actually produced ( $90 %$ yield) $= \frac{7.56 \times 90}{100}$ $= 6.8 = x$ mol of $C l O_{2}$
Hence, the value of $10 x$ is $68$ .

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