What mass of zinc fluoride is formed by the interaction of zinc weighing 16.5 g with an excess of fluorine?
July 31, 2021
m (Zn) = 16.5 g
m (ZnF2) -?
1) Write the equation of a chemical reaction:
Zn + F2 => ZnF2;
2) Calculate the molar masses of Zn and ZnF2:
M (Zn) = Mr (Zn) = Ar (Zn) = 65 g / mol;
M (ZnF2) = Mr (ZnF2) = Ar (Zn) * N (Zn) + Ar (F) * N (F) = 65 * 1 + 19 * 2 = 103 g / mol;
3) Calculate the amount of substance Zn:
n (Zn) = m (Zn) / M (Zn) = 16.5 / 65 = 0.25 mol;
4) Determine the amount of ZnF2 substance:
n (ZnF2) = n (Zn) = 0.25 mol;
5) Calculate the mass of ZnF2:
m (ZnF2) = n (ZnF2) * M (ZnF2) = 0.25 * 103 = 25.8 g.
Answer: The mass of ZnF2 is 25.8 g.
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