Q2 / 60 · KVPY 2015 (SA) · Chemistry
When $262 \mathrm{~g}$ of xenon (atomic mass $=131$ ) reacted completely with $152 \mathrm{~g}$ of fluorine (atomic mass $=19$ ), a mixture of $\mathrm{XeF}_{2}$ and $\mathrm{XeF}_{6}$ was produced. The molar ratio $\mathrm{XeF}_{2}: \mathrm{XeF}_{6}$ is :
- A1: 2
- B1: 4
- Correct answerC$\mathrm1: 1$
- D1: 3
Solution
$\mathrm{Xe}+\mathrm{F}_{2} \rightarrow \mathrm{XeF}_{2}+\mathrm{XeF}_{6}$ $\frac{262}{131} \quad \frac{152}{38}$ a mole $\mathrm{b}$ mole $=2$ mole 4 mole Let a mole $\mathrm{XeF}_{2}$ form and $\mathrm{b}$ mole $\mathrm{XeF}_{6}$ form Apply POAC $$ \begin{array}{ll}\mathrm{a} \times 1+\mathrm{b} \times 1=2 \\ 2 \mathrm{a}+6 \mathrm{~b}=8\end{array} $$ After solving eq. $(1) \&(2)$ $\mathrm{a}=1$ mole $\& \mathrm{~b}=1 \mathrm{~mole}$
Hence, the correct answer is option: 'C'.