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.
From the question, the metal is the limiting reactant and must be used to obtain the required results which are;
a) There are 1.20 mol F in MF2.
b) There are 40g of M in MF2.
c) M is calcium.
From the reaction equation;
M + F2 ----> MF2
Since the reaction is 1:1, 0.600 mol of MF2 is formed
Note that 1 mole of MF2 contains two moles of F and one mole of M
Hence, number of moles of F in MF2 = 2 * 0.600 mol = 1.20 mol F
Since 1 mole of M is contained in MF2, there are 0.600 moles of M in MF2.
If 1 mol of M forms 1 mole of MF2
And, 0.600 mol of M form 0.600 mol of MF2
Number of moles of MF2 = mass/molar mass
Number of moles of MF2 = 0.600 mol
Mass of MF2 formed = 46.8 g
Molar mass of MF2 = mass/Number of moles of MF2
Molar mass of MF2 = 46.8 g/0.600 mol
Molar mass of MF2 = 78 g/mol
Molar mass = Sum of Relative atomic mass of component atoms.
Hence;
78 = M + 2(19)
M = 78 - 2(19)
M = 40 g
The element must be calcium
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