11 moles of $F e S O_{4}$ are treated with one mole of acidic $K M n O_{4}$ solution and resulting solution is further treated with acidic $K_{2} C r_{2} O_{7}$ solution. As a result, remaining $F e S O_{4}$ is completely oxidised. Find out minimum number of moles of $K_{2} C r_{2} O_{7}$ present in the solution.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

${Fe}^{2 +} + {KMnO}_{4} \overset{H^{+}}{⟶} {Mn}^{2 +} + {Fe}^{3 +}$

$11 m o l e s 1 m o l e$

$(e x c e s s r e a g e n t) (L i m i t i n g r e a g e n t)$

$\begin{array}{l} n_{{Fe}^{2 +}} (reacted) \times 1 = n_{{KMnO}_{4}} \times 5 mole \\ \Rightarrow n_{{Fe}^{2 +}} (reacted) = 1 \times 5 = 5 mole \\ So, n_{{Fe}^{2 +}} (left) = 11 - 5 = 6 mole \end{array}$

$Now, reaction with acidic K_{2} {Cr}_{2} O_{7}$

$\begin{array}{l} {Cr}_{2} O_{7}^{2 -} + {Fe}^{2 +} \overset{H^{+}}{⟶} {Cr}^{3 +} + {Fe}^{3 +} \\ x mole 6 mole \\ \Rightarrow x \times 6 = 6 \times 1 \\ ∴ x = 1 \end{array}$