Answer: A mass of 84.46 g of cesium fluoride would be required to make 73.1 g of cesium xenon heptafluoride.
Explanation:
Given: Mass of cesium xenon heptafluoride = 73.1 g
The molar mass of cesium xenon heptafluoride is 131.3 g/mol. So, moles of [tex]CsXeF_{7}[/tex] is calculated as follows.
[tex]Moles = \frac{mass}{molar mass}\\= \frac{73.1 g}{131.3 g/mol}\\= 0.556 mol[/tex]
According to the given equation, 1 mole of CsF yields 1 mole of [tex]CsXeF_{7}[/tex].
Hence, moles of CsF reacting will also be equal to 0.556 mol. As molar mass of CsF is 151.9 g/mol so its mass is calculated as follows.
[tex]Moles = \frac{mass}{molarmass}\\0.556 mol = \frac{mass}{151.9 g/mol}\\mass = 84.46 g[/tex]
Thus, we can conclude that a mass of 84.46 g of cesium fluoride would be required to make 73.1 g of cesium xenon heptafluoride.
