Question

$6.13$ g sample of $K_{2}C{r}_2{O}_7$ ($96\%$ purity) is allowed to react with $320$ mL of $HCl$ solution having density $1.15$ g/mL and containing $30\%$ by mass of $HCl$. What mass of $C{l}_2$ is liberated?
$K_{2}C{r}_2{O}_7+14HCl⟶2KCl+3CrC{l}_3+7H_{2}O+3C{l}_2$

• A. $4.7$ g
• B. $4.3$ g
• C. $7.3$ g
• D. None of these

Answer 1

Answer: (B)

$4.3$ g

$m_{HCl}=\frac{30}{36.5\times \frac{100}{1.15\times 1000}}=9.45M$
Millimoles of $HCl=9.45\times 320=3.24$
Millimoles of $K_{2}C{r}_2{O}_7=6.13\times \frac{96}{100}\times \frac{1000}{294}=20.02$
$\therefore$ Millimoles of $C{l}_2$ formed $=20.03\times 3$
$\therefore w_{C{l}_2}=\frac{20.03\times 3\times 71}{1000}=4.26$ g
Hence, mass of $C{l}_2$ liberated is $4.26$ g.