Calculate the mass of the precipitate formed when 2.27L of 0.0930 M Ba(OH)2 are

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Explanation / Answer

Ba(OH)2+Na2SO4 --> BaSO4 + 2 NaOH no. of moles = molarity*volume So no. of moles Ba(OH)2 = 2.27*0.093 mol = 0.2111 No. of moles of Na2SO4 = 3.23*0.0664 = 0.21442 We need one mol Na2SO4 for one mol Ba(OH)2 So for 0.21111 mol Ba(OH)2 we need 0.21111 mol Na2SO4(remainingNa2SO4 will remain unreacted) This will form 0.21111 mol BaSO4 precipitate Calculate mass of 0.21111 mol BaSO4 now..(I dont remember mass ofBa)

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