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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 :

Maacastrobridn

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
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