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Chemistry, 09.10.2019 16:20 fonsworth5
If a solution containing 46.26 g of mercury(ii) nitrate is allowed to react completely with a solution containing 13.180 g of sodium sulfate according to the equation below. hg(no3)2(aq)+na2so4(aq)⟶2nano3(aq)+ hgso4(s) hg ( no 3 ) 2 ( aq ) + na 2 so 4 ( aq ) ⟶ 2 nano 3 ( aq ) + hgso 4 ( s ) how many grams of solid precipitate will be formed? how many grams of the reactant in excess will remain after the reaction?
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