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3 years ago

A sample of 0.600 mol of a metal M reacts completely with excess fluorine to form 46.8 g of MF2.?

Chemistry

1 answer:

harkovskaia [24]3 years ago

4 0

Answer:

See Explanation

Explanation:

Given:

moles M = 0.600 mole

moles F = excess (for rxn stoichiometry)

Formula Weight (F.Wt.) of F = 19 grams/mole (from Periodic Table)

Yield in grams = 46.8 grams (assuming theoritical yield)

Rxn: M + F₂ => MF₂

0.600mol Excess 0.600mol (1:1 rxn ratio for M:MF₂)

a. moles of F in MF₂ = 2(0.600) moles F = 1.2 moles F

b. mole weight MF₂ = 46.8g/0.600mol = 78g/mole

F.Wt. MF₂ = F.Wt. M + 2(F.Wt. F)

=> mass M = F.Wt. M = [F.Wt. MF₂ - 2(F.Wt. F)] = 78g/mol. - 2(19g/mol.)

= (78 - 38) grams/mole = 40 grams/mole

c. Calcium (Ca) has F.Wt. = 40 grams/mole (compared to Calcium on Periodic Table.

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