Get insightful responses to your questions quickly and easily on IDNLearn.com. Join our knowledgeable community and get detailed, reliable answers to all your questions.

Using the balanced equation below, how many grams of cesium fluoride would be required to make 73.1 g of cesium xenon heptafluoride?

CSF + XeF6 → CsXeF7


Sagot :

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.