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A 10.00 g sample of a soluble barium salt is treated with
an excess of sodium sulfate to precipitate 11.21 g BaSO4
(M = 233.4). Which barium salt is it?
(A) BaCl2 (M = 208.2)
(B) Ba(O2CH)2 (M = 227.3)
(C) Ba(NO3)2 (M = 261.3)
(D) BaBr2 (M = 297.1)

Respuesta :

(11.21 g BaSO4) / (233.4 g/mol BaSO4) = 0.0480 mol BaSO4 and original barium salt 

(10.0 g) / (0.0480 mol) = 208.3 g/mol 
So it must have been BaCl2.

I think it letter A. 

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