Question

Chlorine dioxide $\left(Cl{O}_2\right)$ is used nowadays for water treatment rather than $C{l}_2$. $Cl{O}_2$ is obtained by passing $C{l}_2\left(g\right)$ into a concentrated solution of sodium chlorite $\left(NaCl{O}_2\right)$. The reaction gives $90\%$ yield. If $x$ moles of $Cl{O}_2$ is produced in $3.78$ L of $2.0M$ $NaCl{O}_2\left(aq\right)$, then calculate the value of $10x$.

• A. $68$
• B. $60$
• C. $74$
• D. None of the above

Answer 1

The correct option is A $68$

$\underset{2mol}{2NaCl{O}_2}+C{l}_2\to 2NaCl+\underset{2mol}{2Cl{O}_2}$
Moles of $Cl{O}_2$ $=$ Moles of $NaCl{O}_2$ $=$ $M\times V_L=2\times 3.78$ $=7.56$
Moles of $Cl{O}_2$ actually produced ($90\%$ yield) $=\frac{7.56\times 90}{100}$ $=6.8=x$ mol of $Cl{O}_2$

Hence, the value of $10x$ is $68$.