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Standard XII

Chemistry

Representation of Cell

If 10 g of ...

Question

If $10$ g of $V_{2} O_{5}$ is dissolved in acid and is reduced to $V^{2 +}$ by zinc metal, how many moles of $I_{2}$ could be reduced by the resulting solution if it is further oxidised to $V O^{2 +}$ ions?

0.11

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0.22

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0.055

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0.44

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Solution

The correct option is A $0.11$
Moles of $V_{2} O_{5} = \frac{10}{51 \times 2 + 5 \times 16} = \frac{10}{102 + 80} = \frac{10}{182} = 0.055$ mol

Moles of $V^{+ 2} = 0.055 \times 2$ $= 0.1098$ mol $\approx 0.11$ mol
$V^{+ 2} ⟶ V O^{+ 2} + 2 e^{-}$
$I_{2} + 2 e^{-} ⟶ 2 I^{-}$
Moles of $I_{2}$ $=$ Moles of $V^{+ 2} = 0.11$ mol

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Similar questions

Q. If

10

g of

V_{2} O_{5}

is dissolved in acid and is reduced to

V^{2 +}

by zinc metal, how many moles of

I_{2}

could be reduced by the resulting solution if it is further oxidized to

V O^{2 +}

ions? (Atomic mass of

V

51

g/mol)

Q. If 10 g of

V_{2} O_{5}

is dissolved in acid and is reduced to

V^{2 +}

by zinc metal, how many moles

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, O = 16, I = 127)

10.0

g pure

V_{2} O_{5}

is brought in solution state by dissolving it in acid medium. Now, the solution is reduced to

V^{2 +}

by quantitative mass

Z n

metal. If

I_{2}

is now added to this solution, how much moles of

I_{2}

will be reduced if

V^{2 +}

is oxidized to

V^{4 +}

ions?
Given that atomic mass of

V = 51

Q. In an experiment,

20

g of vanadium (

V

) oxide (molar mass

= 182

) was reduced by excess of zinc dust in acidic solution to vanadium (

I I

) ions. The required number of moles of iodine to reoxidise vanadium (

I I

) to

V O^{2 +}

is :

Q. In an experiment,

20

g of vanadium (

V

) oxide (molar mass

= 182

g/mol) was reduced by excess of zinc dust in acidic solution to vanadium (II) ions. If the required number of moles of iodine to reoxidise vanadium (II) to

V O^{2 +}

22 \times 10^{- x}

, then what is the value of

x

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If 10 g of V2O5 is dissolved in acid and is reduced to V2+ by zinc metal, how many moles of I2 could be reduced by the resulting solution if it is further oxidised to VO2+ ions?

The correct option is A 0.11Moles of V2O5=1051×2+5×16=10102+80=10182=0.055 molMoles of V+2=0.055×2 =0.1098 mol ≈0.11molV+2⟶VO+2+2e−I2+2e−⟶2I−Moles of I2 = Moles of V+2=0.11 mol

If $10$ g of $V_{2} O_{5}$ is dissolved in acid and is reduced to $V^{2 +}$ by zinc metal, how many moles of $I_{2}$ could be reduced by the resulting solution if it is further oxidised to $V O^{2 +}$ ions?