Question

A $Xe{F_6}$ fluorinates ${I_2}$to $I{F_7}$ and liberates Xenon (g). $210{\text{ mmol}}$ of $Xe{F_6}$ can yield a maximum of ________ mmol of $I{F_7}$
A.420
B.180
C.210
D.245

Answer 1

Hint:To solve this question, you must recall basic stoichiometric fundamentals. If we know the amount of reactants in a given reaction, then we can determine the amount of products formed in the reaction.

Complete step by step answer:
The balanced fluorination reaction of iodine to iodine hepta- fluoride using xenon hexa- fluoride can be written as:
$7Xe{F_6} + 3{I_2} \to 7Xe + 6I{F_7}$
From this reaction, we can see that 7 moles of $Xe{F_6}$ on reaction with 3 moles of yield 6 moles of $I{F_7}$. Given the amount of xenon hexa- fluoride in the question, we can easily find the amount of iodine hepta- fluoride produced.
We know that 7 moles of $Xe{F_6}$ gives 6 millimoles of $I{F_7}$.
So one mole of $Xe{F_6}$ will give $ = \dfrac{6}{7}$millimoles of $I{F_7}$
Hence, $210{\text{ mmol}}$ of $Xe{F_6}$ will give $ = \dfrac{6}{7} \times 210$millimoles of $I{F_7}$
So the amount of $I{F_7}$ produced $ = \dfrac{6}{7} \times 210 = 180{\text{ mmols}}$

Thus, the correct answer is B.

Note:
Stoichiometry is based upon the very basic laws of chemistry that help us to understand it better, which are namely the law of conservation of mass, the law of reciprocal proportions, the law of definite proportions (the law of constant composition) and the law of multiple proportions .
In general, different substances combine in definite but fixed ratios in chemical reactions. Since matter can neither be created nor destroyed, nor can one element change into the other in a chemical reaction, thus the amount of each element must be the same throughout the entire reaction. For example, the number of atoms of any element in the reactants will be always equal to the number of atoms of that element in the products formed.