Determine the equivalent weight of $K_{2} C r_{2} O_{7}$ and $F e S O_{4}$ in the reaction.
$K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 F e S O_{4} \to C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 3 F e_{2} (S O_{4})_{3} + 7 H_{2} O$

Open in App

Solution

$K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 F e S O_{4} ⟶ C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 3 F e_{2} (S O_{4})_{3} + 7 H_{2} O$
$K_{2} C r_{2} O_{7} + 6 ⟶ C r_{2} (S O_{4})_{3} + 3$ $n$ -factor= $3 \times 2 = 6$
$2 F e S O_{4} + 2 ⟶ F e_{2} (S O_{4})_{3} + 3$ $n$ -factor= $1$
$∴$ Equivalent weight of $K_{2} C r_{2} O_{7} = \frac{294}{6} = 49$
$∴$ Equivalent weight of $F e S O_{4} = \frac{151.9}{1} = 151.9$

Suggest Corrections

Similar questions

K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 F e S O_{4} \to 3 F e_{2} (S O_{4})_{3} + C r_{2} (S O_{4})_{3} + 7 H_{2} O + K_{2} S O_{4}

6 F e S O_{4} + K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} \to 3 F e_{2} (S O_{4})_{3} + C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 7 H_{2} O

K_{2} C r_{2} O_{7} + 7 H_{2} S O_{4} + 6 F e S O_{4} \to 3 F e_{2} (S O_{4})_{3} + C r_{2} (S O_{4})_{3} + K_{2} S O_{4} + 7 H_{2} O

K_{2} {C r}_{2} O_{7} - -------- - + 7 H_{2} S O_{4} + 6 F e S O_{4} ⟶ K_{2} S O_{4} + {C r}_{2} {(S O_{4})}_{3} + 3 {F e}_{2} {(S O_{4})}_{3} + 7 H_{2} O

M

K_{2} C r_{2} O_{7}

K_{2} C r_{2} O_{7} + 6 K I + 7 H_{2} S O_{4} \to C r_{2} (S O_{4})_{3} + 3 I_{2} + 4 K_{2} S O_{4} + 7 H_{2} O

Join BYJU'S Learning Program