Answer:
Approximately [tex]279\; \rm mg[/tex] of fluorine would be produced.
Explanation:
Look up the relative atomic mass of these calcium and fluorine on a modern periodic table:
In other words, the mass of each mole of calcium atoms is (approximately) [tex]40.078\; \rm g[/tex]. Similarly, the mass of each mole of fluorine atoms is (approximately) [tex]18.998\; \rm g[/tex].
Calculate the number of moles of calcium atom that were produced:
[tex]\begin{aligned} &n({\rm Ca}) \\ &= \frac{m({\rm Ca})}{M({\rm Ca})} \\ &= \frac{294\; \rm mg}{40.078\; \rm g \cdot mol^{-1}} \\ &= \frac{0.294\; \rm mg}{40.08 \; \rm g \cdot mol^{-1}}\\ & \approx 7.3357 \times 10^{-3}\; \rm mol\end{aligned}[/tex].
Every formula unit of calcium fluoride, [tex]\rm CaF_2[/tex], contains twice as many fluorine atoms as calcium atoms.
Hence, if the sample contained approximately [tex]7.33570 \times 10^{-3}\; \rm mol[/tex] calcium atoms, it would contain twice as many fluorine atoms:
[tex]\begin{aligned}n({\rm F}) &= 2\, n({\rm Ca}) \\ &\approx 2 \times 7.33570\times 10^{-3}\; \rm mol \\ &\approx 1.4671 \times 10^{-2}\; \rm mol\end{aligned}[/tex].
Calculate the mass of that many fluorine atoms:
[tex]\begin{aligned}& m({\rm F}) \\ &= n({\rm F}) \cdot M({\rm F}) \\ &\approx 1.4671 \times 10^{-2}\; \rm mol \times 18.998\; \rm g \cdot mol^{-1} \\ &\approx 278.72 \times 10^{-3}\; \rm g \\&\approx 279\; \rm mg\end{aligned}[/tex].
279 mg of Fluorine was formed in the decomposition reaction
The equation of the reaction is;
CaF2 -----> Ca + 2F
Mass of calcium formed = 294 mg
Number of moles of Ca in 294 mg = 294 × [tex]10^-3[/tex] g/40 g/mol = 0.00735 moles
From the equation, the products are in ration 1 : 2
Since 1 mole of Ca is to 2 moles of F
0.00735 moles of Ca is to 0.00735 moles × 2 moles/1 mole
= 0.0147 moles of F
Mass of F = 0.0147 moles × 19 g/mol
= 0.279g or 279 mg of F
Learn more: https://brainly.com/question/9743981
