Double displacement Na2S + Pb(NO3)2 | Sodium sulphide + Lead nitrate | Ionic equation
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When lead (II) nitrate reacts with sodium iodide, sodium nitrate and lead (II) iodide are formed.Balance the following equation:Pb(NO3)2 (aq) 2NaI (aq) → 2NaNO3 (aq) PbI2 (s)If I start with 25 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed?What is the limiting reagent in the reaction described in problem 2?How many grams of lead (II) iodide is formed?How much of the nonlimiting reagent will be left over from the reaction in proble
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