25.4 g of iodine and 14.2 g chlorine react to give a mixture of ICl and `Icl_3`. How many moles of ICl and `"Icl"_(3)` are formed, respectively ?

Options

A. `""_(0.05)^(0.05)`
B. `""_(0.05)^(0.1)`
C. `""_(0.5)^(0.5)`
D. `""_(0.1)^(0.1)`

Answer

D

Explanation

Let x be the number of moles of ICl and y be the number of moles of `"Icl"_(3)` formed.
`I_(2) + 2Cl_(2) to Icl + Icl_(3)`
Mol of `I_(2) = 25.4/254 = 0.1`
Mol. Of `Cl_(2) = 14.2/71 = 0.2`
Apply the principle of conservation of atoms to I,
`0.1 xx 2 = x + 1 + y xx 1`
`0.2 = x+y` .....(1)
Apply the principle of conservation of atoms to Cl,
`0.2 xx 2 = x + 3y = 0.4`
On Solving,
`y = 0.1 `mol
x = 0.1 mol