Question

XeF6 fluorinates I2 to IF7 and liberates Xenon(g). 210 mmol of XeF6 can yield a maximum of_____ mmol of IF7

• A. 420
• B. 180
• C. 210
• D. 245

Answer 1

Answer: (B)

180

Answer 2

(2)

XeF6 + I2 $\longrightarrow$ IF7 + Xe

POAC on ‘F’ :

6 (m.mole of XeF6) = 7 (m.mole of IF7)

$\frac{210 \times 6}{7}$ = m.moles of IF7