Question

how many grams of cesium fluoride would be required to make 73.1 g of cesium xenon heptaflouride?

Answer 1

The mass of cesium fluoride which is required to make 73.1 g of cesium xenon heptaflouride is 27.6 grams.

What is the relation between mass & moles?

Relation between the mass and moles of any substance will be represented as:

n = W/M, where

• W = given mass
• M = molar mass

Moles of CsXeF₇ = 73.1g / 397g/mol = 0.184mol

Given chemical reaction is:

CsF + XeF₆ → CsXeF₇

From the stoichiometry of the reaction, it is clear that:

0.184 moles of CsF required = to react with 0.184 moles of CsXeF₇

Now we convert this moles of CsF into mass as:

W = (0.184mol)(150g/mol) = 27.6 g

Hence required mass of CsF is 27.6g.

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