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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.

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