Question

$25.4\text{g}$ of $I_2$ and $14.2\text{g}$ of $C{l}_2$ are made to react completely to yield a mixture of $ICl$ and $IC{l}_3$. Calculate the number of moles of $ICl$ and $IC{l}_3$. (Given: Molar mass of $I$= $127$ $g/mol$)

• A. $0.2,0.2$
• B. $0.1,0.1$
• C. $0.1,0.3$
• D. $0.3,0.1$

Answer 1

The correct option is B $0.1,0.1$

$I_2+2C{l}_2\to ICl+IC{l}_3$
Number of moles of $I_2=\frac{25.4\text{g}}{254\text{g/mol}}=0.1\text{mol}$
1 mol of $I_2$ produces 1 mol of $ICl$ and 1 mol of $IC{l}_3$
0.1 mol of $I_2$ produces 0.1 mol of $ICl$ and 0.1 mol of $IC{l}_3$