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)

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.

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

the answer is 303.15 or 303