Question

part a how much fluorine was formed? express your answer in grams to three significant figures. the mass ratio of sodium to fluorine in sodium fluoride is 1.21:1. a sample of sodium fluoride produced 55.8 g of sodium upon decomposition.

Explanation:

Step1: Set up mass - ratio equation

Let the mass of sodium be $m_{Na}$ and the mass of fluorine be $m_{F}$. The mass - ratio of sodium to fluorine in sodium fluoride is given as $\frac{m_{Na}}{m_{F}}=1.21$. Also, $m_{Na}+m_{F}=55.8$ g. From $\frac{m_{Na}}{m_{F}} = 1.21$, we have $m_{Na}=1.21m_{F}$.

Step2: Substitute and solve for $m_{F}$

Substitute $m_{Na}=1.21m_{F}$ into $m_{Na}+m_{F}=55.8$ g. So, $1.21m_{F}+m_{F}=55.8$ g. Combining like terms, $(1.21 + 1)m_{F}=55.8$ g, which gives $2.21m_{F}=55.8$ g. Then $m_{F}=\frac{55.8}{2.21}$ g.

Step3: Calculate the value of $m_{F}$

$m_{F}=\frac{55.8}{2.21}\text{ g}\approx25.3$ g

Answer:

$25.3$ g