Question

Then $n$-factor of $HCl$ in the equation
$K_{2}C{r}_2{O}_7+14HCl\to 2KCl+2CrC{l}_3+3C{l}_2+7H_{2}O$ is?

• A. $\frac{2}{3}$
• B. $\frac{11}{3}$
• C. $\frac{7}{3}$
• D. $\frac{3}{7}$

Answer 1

Answer: (D)

$\frac{3}{7}$

According to the Given Balanced Equation

14 Moles of $HCl$ looses 6 Moles of Electrons

1 Moles of $HCl$ looses $\frac{6}{14}$ Moles of Electrons

So N factor is $\frac{3}{7}$