Respuesta :
The mass of fluorine that were formed is 592 mg
Stoichiometry
From the question, we are to determine the mass of fluorine formed
First, we will write the balanced chemical equation for the decomposition reaction
The balanced chemical equation for the reaction is
CaF₂ → Ca + F₂
This means
1 mole of calcium fluoride decomposes to give 1 mole of calcium and 1 mole of fluorine
Now, we will determine the number of moles of calcium produced
From the given information,
Mass of calcium produced = 625 mg = 0.625 g
Using the formula,
[tex]Number\ of \ moles =\frac{Mass}{Atomic\ mass}[/tex]
Atomic mass of calcium = 40.078 g/mol
Then,
Number of moles of calcium produced = [tex]\frac{0.625}{40.078}[/tex]
Number of moles of calcium produced = 0.01559 mole
Since
1 mole of calcium fluoride decomposes to give 1 mole of calcium and 1 mole of fluorine
Then,
0.01559 mole of calcium fluoride will decompose to give 0.01559 mole of calcium and 0.01559 mole of fluorine
∴ Number of mole of fluorine formed was 0.01559 mole
Now, for the mass of fluorine formed
Using the formula,
Mass = Number of moles × Molar mass
Molar mass of fluorine = 38 g/mol
Then,
Mass of fluorine formed = 0.01559 × 38
Mass of fluorine formed = 0.59242 g
Mass of fluorine formed = 592.42 mg
Mass of fluorine formed ≅ 592 mg
Hence, the mass of fluorine that were formed is 592 mg.
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