Nygardtiaidn
Answered

IDNLearn.com: Your trusted platform for finding reliable answers. Find in-depth and accurate answers to all your questions from our knowledgeable and dedicated community members.

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.

We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.