Asked by Susan
A metal M forms the sulphate M2(SO4)3 . A 0.596 gram sample of the sulphate reacts with excess BaCl2 to give 1.220 g BaSO4. What is the atomic weight of M?
(Atomic weights : S = 32, Ba = 137.3)
(Atomic weights : S = 32, Ba = 137.3)
Answers
Answered by
DrBob222
.....M2(SO4)3 + 3BaCl2 ==> 3BaSO4 + 2MCl3
.....0.596g.................1.220g
Convert 1.220g BaSO4 to mols. mol = g/molar mass = estimated 0.005
Convert to mols M2(SO4)3. That's estd 0.005 x (1 mol M2(SO4)3/3 mol BaSO4) = 0.005*1/3 = about 0.00174 but you need to sharpen the numbers.
Then mols = g/molar mass. You know mols and g, solve for molar mass. That's an estimated 342 and the 342 must be 2*M + 3*SO4. You know S and O, solve for M. I obtained approximately 27.
.....0.596g.................1.220g
Convert 1.220g BaSO4 to mols. mol = g/molar mass = estimated 0.005
Convert to mols M2(SO4)3. That's estd 0.005 x (1 mol M2(SO4)3/3 mol BaSO4) = 0.005*1/3 = about 0.00174 but you need to sharpen the numbers.
Then mols = g/molar mass. You know mols and g, solve for molar mass. That's an estimated 342 and the 342 must be 2*M + 3*SO4. You know S and O, solve for M. I obtained approximately 27.
Answered by
D. Teja
It should be clear not understanding
Answered by
Pranshu
Nice question. Try to solve it by yourself. Not that difficult really. Go ahead.