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Ca = 48.1, F = 19.0 -> 2F = 38.0
--> CaF2 = 78.1
%F/CaF2 = 38.0/78.1 = 48.7%
-->
[tex]0.487 \times 145 = 70.615[/tex]

Answer:

The amount of F present in 145 g CaF2 is 35.3 g (≅35 g)

Explanation:

Given:

Mass of CaF2 = 145 g

To determine:

Mass of F present in 145 g of CaF2

Explanation:

Based on the formula stoichiometry for CaF2:

1 mole of Caf2 contains 2 moles of F

Molar mass of CaF2 = Atomic weight of Ca + 2(atomic weight of F)

= 40 + 2(19) = 78 g

Atomic mass of F = 19 g

i.e.  78 g of CaF2 contains 19 g of F

Amount of F present in the given 145 g of CaF2 would be:

[tex]=\frac{145 g\ CaF2 * 19g\ F}{78 g \ CaF2 } =35.3 g\ F[/tex]

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