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A sample of 0.600 mol of a metal M reacts completely with excess fluorine to form 46.8 g of MF2.?
a) How many moles of F are in the sample of MF2 that forms? (1.20 mol F)
b) How many grams of M are in this sample of MF2?
c) What element is represented by the symbol M? (Ca)

Sagot :

drpelezo

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.

pstnonsonjoku

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

Learn more: https://brainly.com/question/17275626

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