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245 g of iodine and 142 g of chlorin...

245 g of iodine and 142 g of chlorine are made to react completely to given a mixture of `ICl` and `ICl_(3)` .How many moles of each are formed ?

A

0.1 mole of IC l and 0.1 mole of `ICI_(3)`

B

1 mole of I Cl and 1 mole of `ICL_(3)`

C

0.5 mole of IC l and 0.1 mole of `ICl_(3)`

D

0.5 mole of IC l and 0.1 mole of IC l mole of `ICl_(3)`

Text Solution

Verified by Experts

The correct Answer is:
B
Both reasctants are completely consumed, hence, both are limiting.
`I_(2) + 2Cl_(2) to ICl+ ICl_(3)`
254g or 1 mole `I_(2)` and 142 g or `Cl_(2)` will react to given 1 mole Icl and 1 mole `IC l_(3)`.]
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25.4 gm of iodine and 14.2 gm of chlorine are made to react completely ot yield mixture of ICI and ICI_(3) Ratio of moles of ICI & ICI_(3) formed is (Atomic mass I: 127, Cl=35.5)

25.4 g of iodine and 14.2 g of chlorine are made to react completely to yield a mixture of ICl and ICI_3 . Calculate the ratio of the moles of ICl and Icl_(3) .

Knowledge Check

  • 25.4 gm of iodine and 14.2 gm of chlorine are made to react completely to yeild a mixture of ICI and ICI_(3) ratio of moles of ICI & ICI_(3) formed is (Atomic mass : I = 127 , Cl = 35.5 )

    A
    `1:1`
    B
    `1:2`
    C
    `1:3`
    D
    `2:3`

  • 25.4g of iodine and 14.2g of chlorine are made to react completely to yield a mixture of ICI and ICI_(3) . Calcualte the number of moles of Icl and Icl_(3) formed.

    A
    0.1 mole, 0.1 mole
    B
    0.1 mole, 0.2 mole
    C
    0.5 mole, 0.5 mole
    D
    0.2 mole, 0.2 mole

  • 25.4 g of iodine and 14.2 g of chlorine are made to react completely to yield a mixture of lCl and lCl_(3) .Calculate the ratio of moles of lCl and lCl_(3) .

    A
    `1:1`
    B
    `1:2`
    C
    `1: 3`
    D
    `2 :3`

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