10-2 mol of Fe 3 O 4 is treated with excess of KI solution in the presence of dilute H 2 SO 4, the products are Fe 2+ and I 2( g ). What volume of 0.1( M ) Na 2 S 2 O 3 will be needed to reduce the liberated I 2( g ) ?

Question

10-2 mol of Fe 3 O 4 is treated with excess of KI solution in the presence of dilute H 2 SO 4, the products are Fe 2+ and I 2( g ). What volume of 0.1( M ) Na 2 S 2 O 3 will be needed to reduce the liberated I 2( g ) ? (A) 50 ml (B) 100 ml (C) 200 ml (D) 400 ml

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/293a02333a411af72766c43b3ee90b94-.png" /> ∴ Fe 3 O 4+2 I- arrow Fe 2++ I 2 ∴ No. of moles of I 2 produced =10-2 mol Let v ml 0.1( M ) Na 2 S 2 O 3 solution is required ∴ v × 10-4=2 × 10-2 or v=200 ml.

Q. $1 0^{- 2} m o l$ of $F e_{3} O_{4}$ is treated with excess of $K I$ solution in the presence of dilute $H_{2} S O_{4}$ , the products are $F e^{2 +}$ and $I_{2} (g)$ . What volume of $0.1 (M) N a_{2} S_{2} O_{3}$ will be needed to reduce the liberated $I_{2} (g)$ ?

$50 m l$

$100 m l$

$200 m l$

$400 m l$

$∴ F e_{3} O_{4} + 2 I^{-} \to F e^{2 +} + I_{2}$
$∴$ No. of moles of $I_{2}$ produced $= 1 0^{- 2} m o l$
Let $v m l 0.1 (M) N a_{2} S_{2} O_{3}$ solution is required
$∴ v \times 1 0^{- 4} = 2 \times 1 0^{- 2}$
or $v = 200 m l$ .

Q. 10−2mol of Fe3​O4​ is treated with excess of KI solution in the presence of dilute H2​SO4​, the products are Fe2+ and I2​(g). What volume of 0.1(M)Na2​S2​O3​ will be needed to reduce the liberated I2​(g) ?

50ml

100ml

200ml

400ml