Question

Calculate the volume of 0.1 M K2Cr2O7 required to oxidize 35mL of 0.5 M Fe2SO4 solution​

Answer 1

Explanation:

$\huge\bf{\mid{\overline{\underline{ANSWER:-}\mid}}}$

$\huge\bold{Given:-} \\ \\ molarity \: \: of \: \: K_{2}Cr_{2}O_{7} \: \: is \: \: 0.1 \: M \\ \\ molarity \: \: of \: \: FeSO_{4} \: \: is \: \: 0.5 \: M \\ \\ Volume \: \: of \: FeSO_{4} \: \: is \: \: 35 \: mL \\ \\ \\ \huge\bold{To find:-} \\ \\ Volume \: \: of \: K_{2}Cr_{2}O_{7}$

The reaction is as follows,

$K_{2}Cr_{2}O_{7} + 7H_{2}SO_{4} + 6FeSO_{4} \\ \longrightarrow K_{2}SO_{4} + Cr_{2}({SO_{4}}) _{3} \\ + 7H_{2}O + 3Fe_{2}({SO_{4}}) _{3}$

$\large\underline{By \: Molarity \: Equation}$

$\frac{M_{1}V_{1}}{n_{1}} \ = \frac{M_{1}V_{1}}{n_{2}} \\ \\ \frac{0.1 \times V_{1}}{1} = \frac{0.5 \times 3.5}{6} \\ \\ \huge \boxed{ \implies V_{1} = 29.2mL}$