If 10 g of V2 O5 is dissolved in acid and is reduced to V2+ by zinc metal, how many mole I2 could be reduced by the resulting solution if it is further oxidised to VO2+ ions? [Assume no change in state of Zn2+ ions] (V = 51,0 = 16, I = 127):

Question

If 10 g of V2 O5 is dissolved in acid and is reduced to V2+ by zinc metal, how many mole I2 could be reduced by the resulting solution if it is further oxidised to VO2+ ions? [Assume no change in state of Zn2+ ions] (V = 51,0 = 16, I = 127): (A) 0.11 mole of I2 (B) 0.22 mole of I2 (C) 0.055 mole of I2 (D) 0.44 mole of I2

Tardigrade Admin · Accepted Answer

Mole of V2 ,O5 ,=(10/51×2×5×16)=(10/102+8) =(10/182) =0.55 Mole of r+2 = .055 × 2 = .1098 mole ≃ 0.11 V+2 → overset+4VO+2 ⇒ Moles of I2 = Moles of V+2 = .11

Q. If 10 g of $V_{2} O_{5}$ is dissolved in acid and is reduced to $V^{2 +}$ by zinc metal, how many mole $I_{2}$ could be reduced by the resulting solution if it is further oxidised to $V O^{2 +}$ ions? [Assume no change in state of $Z n^{2 +}$ ions] ( $V = 51, 0 = 16, I = 127$ ):

$0.11$ mole of $I_{2}$

$0.22$ mole of $I_{2}$

$0.055$ mole of $I_{2}$

$0.44$ mole of $I_{2}$

$M o l e o f V_{2}, O_{5}, = \frac{10}{51 \times 2 \times 5 \times 16} = \frac{10}{102 + 8}$
$= \frac{10}{182} = 0.55$
Mole of $r^{+ 2} = .055 \times 2$
$= .1098 m o l e ≃ 0.11$
$V^{+ 2} \to V + 4 O^{+ 2}$
$\Rightarrow M o l eso f I_{2} = M o l eso f V^{+ 2} = .11$

Q. If 10 g of V2​O5​ is dissolved in acid and is reduced to V2+ by zinc metal, how many mole I2​ could be reduced by the resulting solution if it is further oxidised to VO2+ ions? [Assume no change in state of Zn2+ ions] (V=51,0=16,I=127):

0.11 mole of I2​

0.22 mole of I2​

0.055 mole of I2​

0.44 mole of I2​