The lead nitrate and potassium iodide each dissolve and begin to diffuse through the water. When the lead ions and iodide ions meet they react to form solid yellow lead iodide which precipitates out of solution. lead nitrate supplies the lead ion and the potassium iodide the iodide ion to make the insoluble salt lead iodide, which is a yellow crystalline solid. Both chemicals can cause skin, eye and respiratory irritation. At the Lead (II) Nitrate station, use the funnel and graduated cylinder to measure approximately 10.0 mL a. Pb(aq) + 2I – (aq) → PbI 2 … Materials: Lead (II) Nitrate Potassium Iodide 3 beakers Ring Stand Funnel Filter Paper Scale Water Stirring Rod Funnel clamp Wash Bottle Procedure: The first thing that we did was measure out 3.16g Pb(NO3)2 and 5.00 g of KI on the scale. The balanced chemical reaction : Types of reaction : Double-displacement reaction : It is a type of reaction in which two reactants exchange their ions to form two new compounds. Add 5 to 10 drops of potassium iodide solution to the test tube and record your observations of the reaction. Fill a 50 mL beaker with 25 mL 0.1M potassium iodide. The potassium ion is "K"^+ and the iodide ion is "I"^-. Step 2. Pb (NO 3) 2(aq) + 2 K I (aq) ==> PbI 2 (s) + 2KNO 3(aq) Safety . Lead nitrate and potassium iodide should both be considered hazardous. Measure 10 mL of 0.1M lead nitrate into a 25 mL graduated cylinder. 2- 500 mL containers . Lead nitrate and potassium iodide, which are the reactants, are powders that react to form a yellow-colored lead iodide and a colorless potassium nitrate. The equation tells you that 2 moles of potassium iodide will react with 1 mole of lead nitrate to produce 2 moles of potassium nitrate and 1 mole of lead iodide. lead(II) nitrate + potassium iodide ==> lead(II) iodide + potassium nitrate. Can you describe this as a double decomposition reaction? Prove that two solids can react together, making lead iodide from lead nitrate and potassium iodide This is a very quick demonstration, and should take no longer than two minutes. Next you need to find out how many moles of each of the reactants you started with. Magnesium metal and hydrochloric acid solution Place … As a chemistry demonstration, it involves adding equal amounts of the reactants to a flask, placing a stopper on it, and then shaking the flask until the color change to yellow is observed. Large Bottle of Lead (II) Nitrate Large Bottle of Potassium Iodide Waste Container Graduated Cylinder Erlenmeyer Flask Funnel Steps for completion of the lab: 1. 1000 mL beaker . Pour lead nitrate into the potassium iodide slowly. Lead (II) nitrate and potassium iodide solutions Pour about 2.0 mL of lead (II) nitrate into the test tube. The two charges balance in a 1 : 1 ratio, so potassium iodide is simply "KI". This is because lead nitrate is “slightly soluble.” 0.08 g of it can dissolve in 100 mL of water: the yellow clouds only remain visible when there is more than 80 mg present per 100 mL of water. Each group obtains an Erlenmeyer flask 2. Observations of Products : Lead iodide is in solid state and potassium nitrate in aqueous state. Step Seven: Potassium Iodide and Lead Nitrate (Materials: 25 mL 0.1M potassium iodide, 10 mL 0.1M lead nitrate, 50 mL beaker, and a 25 mL graduated cylinder.) glass stir rod . lead nitrate + potassium iodide → lead iodide + potassium nitrate. The solid precipitate is lead iodide. Lead Nitrate Pb(NO 3) 2. Potassium Iodide KI . To balance the charges, we require two nitrate ions per lead (II) ion, and so lead (II) nitrate is "Pb(NO"_3")"_2 . In lead (II) iodide, the charges balance in a 1 : 2 ratio, so the formula is "PbI"_2. Show students that two solids can react together as white lead nitrate and white potassium iodide react to make yellow lead iodide. 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