Home
Class 11
CHEMISTRY
If 10.0 g V2O5 is dissolbed in acid and ...

If 10.0 g `V_2O_5` is dissolbed in acid and reduced to `V^(2+)` by treatment with tin (Sn) metal how many moles of `I_2` could be reduced by the resulting `V^(2+)` solution as it is oxidised to `V^(4+)`? (Atomic weight of V is 51)

Text Solution

AI Generated Solution

To solve the problem step by step, we will follow these calculations: ### Step 1: Calculate the number of moles of `V2O5` To find the number of moles of `V2O5`, we use the formula: \[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \] ...
Promotional Banner

Topper's Solved these Questions

  • STOICHIOMETRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 3.4 (A)|7 Videos
  • STOICHIOMETRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 3.4 (B)|18 Videos
  • STOICHIOMETRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 3.2|3 Videos
  • STATES OF MATTER

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises (Ture False)|25 Videos
  • THERMODYNAMICS

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives (Subjective)|23 Videos

Similar Questions

Explore conceptually related problems

If 10g 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 , O=16 , I=127 )

How many moles of FeC_2O_4 are required to reduce 2 " mol of " KMnO_4 in acidic medium?

How many Faradays are required to reduce 0.25g of Nb (V) to the metal?

1 g equivalent of V_2O_5 in the given reaction is equal to V_2O_5 + Zn rarr ZnO + V (Glven at wt. of V = A)

1 g equivalent of V_2O_5 in the given reaction is equal to V_2O_5 + Zn → ZnO + V (Given at. wt. of V = A)

o-oxylene when oxidised in presence of V_(2)O_(5) the product is :

50 ml of '20 V' H_(2)O_(2) is mixed with 200 ml, '10 V' H_(2)O_(2). The volume strength of resulting solution is:

V_2O_5 reacts with alkalies as well as acids to give

100mL of H_2O_2 is oxidised by 100mL of 0.01M KMnO_4 in acidic medium (MnO_4^(ɵ) reduced to Mn^(2+) ). 100mL of the same H_2O_2 is oxidised by VmL of 0.01M KMnO_4 in basic medium. Hence V is

100mL of H_2O_2 is oxidised by 100mL of 0.01M KMnO_4 in acidic medium (MnO_4^(ɵ) reduced to Mn^(2+) ). 100mL of the same H_2O_2 is oxidised by VmL of 0.01M KMnO_4 in basic medium. Hence V is