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Answer :
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]
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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]
