A 1.103 g sample of sodium fluoride is dissolved in water, and then a precipitate of calcium fluoride is produced by adding a calcium nitrate solution. If the dried calcium fluoride precipitate has a mass of 0.947 g, what is the percent yield?

The percent yield of the reaction is, 92.0 %

Explanation : Given,

Mass of = 1.103 g

Molar mass of = 42 g/mol

Molar mass of = 78 g/mol

First we have to calculate the moles of

Now we have to calculate the moles of

The balanced chemical equation is:

From the reaction, we conclude that

As, 2 mole of react to give 1 mole of

So, 0.0263 mole of react to give mole of

Now we have to calculate the mass of

Now we have to calculate the percent yield of the reaction.

Experimental yield = 0.947 g

Theoretical yield = 1.029 g

Now put all the given values in this formula, we get:

Therefore, the percent yield of the reaction is, 92.0 %