Question

The equivalent weight of $HCl$ in the given reaction is:
$K_{2}C{r}_2{O}_7+14HCl\to 2KCl+2CrC{l}_3+3C{l}_2+H_{2}O$

• A. 16.25
• B. 36.5
• C. 73
• D. 85.17

Answer 1

The correct option is D 85.17

$14\stackrel{1+}{H}\stackrel{1-}{Cl}\to \stackrel{1+}{K}\stackrel{1-}{Cl}+2\stackrel{3+}{Cr}{\stackrel{1-}{Cl}}_3+3{\stackrel{0}{Cl}}_2+{\stackrel{1+}{H}}_2\stackrel{2-}{O}$
$n_f=\frac{6(0-(-1\left)\right)}{14}=6/14$
$Eq{l}^{t}weight=\frac{molecularweight}{n_f}=85.17$