excess sodium reacts with 1.40 moles of fluorine gas according to the equation below. the reaction generates…

Explanation:

Step1: Determine moles of NaF produced theoretically

From the balanced - chemical equation $2Na + F_2\rightarrow2NaF$, the mole ratio of $F_2$ to $NaF$ is $1:2$. Given $n(F_2)=1.40$ moles. So, $n_{theo}(NaF) = 2\times n(F_2)$. $n_{theo}(NaF)=2\times1.40\ mol = 2.80\ mol$

Step2: Calculate the theoretical mass of NaF

The molar mass of $NaF$ is $M(NaF)=22.99\ g/mol + 19.00\ g/mol=41.99\ g/mol$. Using the formula $m = n\times M$, the theoretical mass $m_{theo}(NaF)=n_{theo}(NaF)\times M(NaF)$. $m_{theo}(NaF)=2.80\ mol\times41.99\ g/mol\approx117.57\ g$

Step3: Calculate the percent - yield

The percent - yield formula is $\text{Percent yield}=\frac{m_{actual}}{m_{theo}}\times100%$. Given $m_{actual}=25.9\ g$ and $m_{theo}\approx117.57\ g$. $\text{Percent yield}=\frac{25.9\ g}{117.57\ g}\times100%\approx22.0%$

Answer:

$22.0$