25.4 g of iodine and 7.1 g chlorine are made to react to give a mixture of ICl and $I C l_{3}$ . Number of moles of ICl and $I C l_{3}$ formed are respectively ?

Q. 254 g of iodine and 142g of chlorine are made to react completely to give a mixture of

I C l

and

I C l_{3}

. The moles of each one formed is:

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25.4 g of iodine and 7.1 g chlorine are made to react to give a mixture of ICl and ICl3. Number of moles of ICl and ICl3 formed are respectively ?

The correct option is D 0.05 & 0.05 I2+2Cl2→ICl+ICl3 Mol of I2=25.4254=0.1;Mol of Cl2=7.171=0.1 Chloring is limiting reactant. Hence, number of mol of ICl = number of mol of ICl3=0.12=0.05

25.4 g of iodine and 7.1 g chlorine are made to react to give a mixture of ICl and $I C l_{3}$ . Number of moles of ICl and $I C l_{3}$ formed are respectively ?

The correct option is D
0.05 & 0.05

$I_{2} + 2 C l_{2} \to I C l + I C l_{3}$
Mol of $I_{2} = \frac{25.4}{254} = 0.1; M o l o f C l_{2} = \frac{7.1}{71} = 0.1$
Chloring is limiting reactant. Hence, number of mol of
ICl = number of mol of $I C l_{3} = \frac{0.1}{2} = 0.05$