$Xe{F}_6$ fluorinates $I_2\mathit{ }to\mathit{ }I{F}_7$ and liberates Xenon $\left(g\right).210m\mathit{ }mol\mathit{ }of\mathit{ }Xe{F}_6$. can yield a maximum of ... $mmol\mathit{ }of\mathit{ }I{F}_7$.

The balanced chemical equation is $7{\mathrm{XeF}}_6+3{\mathrm{I}}_2\to 7\mathrm{Xe}+6{\mathrm{IF}}_7$

$7\text{ moles of }{\mathrm{XeF}}_6\text{ on reaction gives }6\text{ moles of IF . }$

$\text{ So, }210\text{ mmoles of }{\mathrm{XeF}}_6\text{ on reaction will give }=\frac{6}{7}\times 210=180\text{ moles of }{\mathrm{IF}}_7\text{. }$