Question

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

Answer 1

Answer:

27.9 g

Explanation:

CsF + XeF₆ → CsXeF₇

First we convert 73.1 g of cesium xenon heptafluoride (CsXeF₇) into moles, using its molar mass:

• Molar mass of CsXeF₇ = 397.193 g/mol

• 73.1 g CsXeF₇ ÷ 397.193 g/mol = 0.184 mol CsXeF₇

As 1 mol of cesium fluoride (CsF) produces 1 mol of CsXeF₇, in order to produce 0.184 moles of CsXeF₇ we would need 0.184 moles of CsF.

Now we convert 0.184 moles of CsF to moles, using the molar mass of CsF:

• Molar mass of CsF = 151.9 g/mol

• 0.184 mol * 151.9 g/mol = 27.9 g