25.4 gm of iodine and 14.2 gm of chlorin...

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)

Text Solution

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According to given data
`underset("1 mol")(I_(2)) + underset("2 mol")(2Cl_(2)) to underset("1 mol")("Icl") + underset("1 mol")(Icl_(3))`
1 mol of `I_(2) = 254 g`
Moles of `I_2` available = `25.4/254 = 0.1` mol
Moles of `Cl_2` available `= 14.2/71 = 0.2` mol
1 mol of `I_2` yields = 1 mole ICl
`therefore` 0.1 mol of `I_2` yields = 0.1 mol ICl And 1 mol of `I_2` yields = 1 mol `ICl_3`
`therefore` 0.1 mol of `I_2` yields = 0.1 mol `ICl_3`
`therefore` Ratio of the moles of ICl and `ICl_3 = 0.1/0.1 = 1:1`

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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 )

`1:1`

`1:2`

`1:3`

`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.

0.1 mole, 0.1 mole

0.1 mole, 0.2 mole

0.5 mole, 0.5 mole

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) .

`1:1`

`1:2`

`1: 3`

`2 :3`