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A chemist added an excess of sodium sulfate to a solution of a soluble barium compound to precipitate all of the barium ion barium sulfate, BaSO4. How many grams of barium ion are in a 403 mg sample of the precipitate? What is the mass percentage of barium in the compound? So far, I have the balanced equation: Na2+SO4 + Ba = BaSO4 + 2Na
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there is 137g Ba in 233g BaSO4. thus, 403g of BaSO$ will have (137/233)*(403) the answer comes 236.8g....
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