15. In the following reaction, 6.00 kg of CaF2 is treated with an excess of H2SO

Question

15. In the following reaction, 6.00 kg of CaF2 is treated with an excess of H2SO4, and produces 2.86 kg of HF. Calculate the percent yield of HF CaF2 + H2SO4 CaSO4 +2HF A) 93.0% B) 30.8% C) 66.0% D) 33.0% E) 48.8% 16. A metal, M, is converted to the metal sulfate, M2(S04)3. Then a solution of the metal sulfate was treated with calcium chloride to give calcium sulfate precipitate. If 1.200g of the metal gave 5.451 g calcium sulfate, what is the identity of metal M? M2(SOJ3(aq) +3CaCl2(aq) 2MCI3(aq) +3CaSO4(s) A) Sc B) Al C) Cr D) Mn E) Co Page 4

Explanation / Answer

Step 1: calculate mol of CaSO4

Molar mass of Ba(OH)2,
MM = 1*MM(Ba) + 2*MM(O) + 2*MM(H)
= 1*137.3 + 2*16.0 + 2*1.008
= 171.316 g/mol


mass(Ba(OH)2)= 4.33 mg
= 0.00433 g

number of mol of Ba(OH)2,
n = mass of Ba(OH)2/molar mass of Ba(OH)2
=(0.00433 g)/(171.316 g/mol)
= 2.527*10^-5 mol
volume , V = 1 L


Molarity,
M = number of mol / volume in L
= 2.527*10^-5/1
= 2.527*10^-5 M

Molar mass of CaSO4,
MM = 1*MM(Ca) + 1*MM(S) + 4*MM(O)
= 1*40.08 + 1*32.07 + 4*16.0
= 136.15 g/mol


mass(CaSO4)= 5.451 g

number of mol of CaSO4,
n = mass of CaSO4/molar mass of CaSO4
=(5.451 g)/(136.15 g/mol)
= 4.004*10^-2 mol

step 2: calculate mol of M2(SO4)3
using reaction,
mol of M2(SO4)3 reacted = (1/3)*moles of CaSO4
= (1/3)*4.004*10^-2 mol
= 0.01335 mol

step 3: calculate mol of M
In 1 mol of M2(SO4)3, there are 2 moles of M
So,
moles of M = 2* mol of M2(SO4)3
= 2*0.01335 mol
= 0.0267 mol

step 4: calculate molar mass of M
use:
mol of M = mass / molar mass
0.0267 mol = 1.200 g/MM
MM = 44.94 g/mol

This is molar mass of Scandium
Answer: A

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