Respuesta :
Answer: The percent yield of the reaction is, 92.0 %
Explanation : Given,
Mass of [tex]NaF[/tex] = 1.103 g
Molar mass of [tex]NaF[/tex] = 42 g/mol
Molar mass of [tex]CaF_2[/tex] = 78 g/mol
First we have to calculate the moles of [tex]NaF[/tex]
[tex]\text{Moles of }NaF=\frac{\text{Given mass }NaF}{\text{Molar mass }NaF}[/tex]
[tex]\text{Moles of }NaF}=\frac{1.103g}{42g/mol}=0.0263mol[/tex]
Now we have to calculate the moles of [tex]CaCl_2[/tex]
The balanced chemical equation is:
[tex]2NaF+Ca(NO_3)_2\rightarrow CaF_2+2NaNO_3[/tex]
From the reaction, we conclude that
As, 2 mole of [tex]NaF[/tex] react to give 1 mole of [tex]CaF_2[/tex]
So, 0.0263 mole of [tex]NaF[/tex] react to give [tex]\frac{0.0263}{2}=0.0132[/tex] mole of [tex]CaF_2[/tex]
Now we have to calculate the mass of [tex]CaF_2[/tex]
[tex]\text{ Mass of }CaF_2=\text{ Moles of }CaF_2\times \text{ Molar mass of }CaF_2[/tex]
[tex]\text{ Mass of }CaF_2=(0.0132moles)\times (78g/mole)=1.029g[/tex]
Now we have to calculate the percent yield of the reaction.
[tex]\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Experimental yield = 0.947 g
Theoretical yield = 1.029 g
Now put all the given values in this formula, we get:
[tex]\text{Percent yield}=\frac{0.947g}{1.029g}\times 100=92.0\%[/tex]
Therefore, the percent yield of the reaction is, 92.0 %
