Question

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

a) 420 b) 180 c) 210 d) 245

Answer 1

7XeF6 + 3I2 ➡️ 6IF7 + 7Xe.

So,

7 mol Xef6 ➡️ 6IF7

So, 210 mmole ➡️ x

So, x = 180 mmole.

Option (b).

Answer 2

Answer:  b) 180

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

$7XeF_6+3I_2\rightarrow 6IF_7+7Xe$

According to stochiometry:

7 moles of $XeF_6$ gives 6 moles of $IF_7$   $(1millimole=10^{-3}moles)$

0.007 milli moles of $XeF_6$ gives 0.006 milli moles of $IF_7$

210 milli moles of $XeF_6$ gives =$\frac{0.006}{0.007}\times 210=180$ milli moles of $IF_7$