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M(BaSO₄) = 137g/mol + 32g/mol + 4*16g/mol = 233g/mol
BaCl₂ + K₂SO₄ → BaSO₄ + 2KCl
208g-----------------233g
20,8g-----------------x
X= 20,8g*233g/208g = 23,3g
Otrzymamy 23,3g siarczanu(VI) baru