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When a solution containing 1.4g of Ba(NO3)2 and 2.4 g of NH2SO3H is boiled, a precipitate forms. This precipitate may be Ba(NH2SO3)2, BaSO4, or BA(NH2)2. You can find these products in the reactions whose equations you balanced in the first reaction.CALCULATE THE NUMBER OF MOLES OF EACH REACTANT
Ba(NO3)2+NH2SO3H+H2O----BaSO4+HNO3+NH4NH3
Ba(NO3)2+NH2SO3H+H2O-----BaSO4+NH4NO3+HNO3
Ba(NO3)2+2NH2SO3H+2H2O------Ba(NH2)2+2H2SO4+2HNO3
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