Question

25.4g of iodine and 14.2g of chlorine are made to react completely to yield a mixture of ICl and ICl₃. Calculate the number of moles of ICI and ICl₃ formed.

Answer 1

The relevant balanced chemical equation is

\(\underset{\;\\1 \text{mole}\\2 \times 127\\=254g}{I_2} + \underset{\;\\2\text{moles}\\2(2\times 35.5)\\= 142 g}{2Cl_2} \longrightarrow \underset{\;\\1\text{mole}}{ICl} + \underset{\;\\1\text{mole}}{ICl_3}\)

According to equation, 254 g of I₂ form = 1 mole of ICl 

25.4 g of 12 form = \(\frac1{254}\) x 25.4 mole of ICI = 0.1 mole of ICI

Similarly, 254 g of I₂ form = 1 mole of ICl₃

1 25.4 g of I₂ form = \(\frac1{254}\) x 25.4 mole of ICl₃ = 0.1 mole of ICl₃