Question

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 $VO^{2+}$ ions? [Assume no change in state of $Zn^{2+}$ ions] ($V = 51,0 = 16, I = 127$):

• A. $0.11$ mole of $I_2$
• B. $0.22$ mole of $I_2$
• C. $0.055$ mole of $I_2$
• D. $0.44$ mole of $I_2$

Answer 1

Answer: (A)

$Mole \,of\, V_{2} ,O_{5} ,=\frac{10}{51\times2\times5\times16}=\frac{10}{102+8}$
$=\frac{10}{182} =0.55$
Mole of $r^{+2} = .055 \times 2 $
$= .1098 mole \simeq 0.11$
$V^{+2} \to \overset{+4}{V}O^{+2} $
$\Rightarrow Moles of I_{2} = Moles of V^{+2} = .11$