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Solution Stoichiometry

Solution Stoichiometry. Beauregard combined 100. mL of 0.150 M potassium iodide and 350. mL of 0.100 M lead II nitrate. A yellow precipitate of lead II iodide is formed in the reaction. Beauregard combined 100. mL of 0.150 M potassium iodide and

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Solution Stoichiometry

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  1. Solution Stoichiometry Beauregard combined 100. mL of 0.150 M potassium iodide and 350. mL of 0.100 M lead II nitrate. A yellow precipitate of lead II iodide is formed in the reaction.

  2. Beauregard combined 100. mL of 0.150 M potassium iodide and 350. mL of 0.100 M lead II nitrate. A yellow precipitate of lead II iodide is formed in the reaction. a) Write a balanced equation for the reaction. The reaction is a double replacement reaction. 2 KI + Pb(NO3)2 PbI2 + 2 KNO3

  3. Beauregard combined 100. mL of 0.150 M potassium iodide and 350. mL of 0.100 M lead II nitrate. A yellow precipitate of lead II iodide is formed in the reaction. 2 KI + Pb(NO3)2 PbI2 + 2 KNO3 b) What is the balanced ionic equation for the reaction? Since the reactants are soluble ionic compounds, they are ionized when dissolved in water. However, the precipitate PbI2 is a solid and no longer in solution. 2 K + + 2 I – + Pb 2+ + 2 NO3 –  2 K + + 2 NO3 – + PbI2 The possible ions present in solution at the end of the reaction are: K +, NO3– ,and any remaining Pb 2+ , I –

  4. Beauregard combined 100. mL of 0.150 M potassium iodide and 350. mL of 0.100 M lead II nitrate. A yellow precipitate of lead II iodide is formed in the reaction. 2 KI + Pb(NO3)2 PbI2 + 2 KNO3 c) What is the theoretical yield of precipitate in grams and what is the limiting reactant? The precipitate is PbI2. 100. mL KI 0.150 M KI1 mol PbI2461 g = 3.46 g PbI2 1000 mL 2 mol KI 1 mol 350. mL Pb(NO3)2 0.1 M Pb(NO3)21 mol PbI2461 g = 16.1 g PbI2 1000 mL 1 mol Pb(NO3)2 1 mol The theoretical yield is the smaller yield – 3.46 g PbI2 The limiting reactant is therefore KI

  5. Beauregard combined 100. mL of 0.150 M potassium iodide and 350. mL of 0.100 M lead II nitrate. A yellow precipitate of lead II iodide is formed in the reaction. 2 KI + Pb(NO3)2 PbI2 + 2 KNO3 d) What is the concentration of potassium ions and iodide ions at the end of the reaction? i) The precipitate is PbI2 and KI is the limiting reactant so the concentration of I – is approximately 0 M

  6. Beauregard combined 100. mL of 0.150 M potassium iodide and 350. mL of 0.100 M lead II nitrate. A yellow precipitate of lead II iodide is formed in the reaction. 2 KI + Pb(NO3)2 PbI2 + 2 KNO3 d) What is the concentration of potassium ions and iodide ions at the end of the reaction? ii) None of the potassium ions precipitated so the amount of potassium does not change but volume changes must be accounted for: 350. mL of Pb(NO3)2 solution were added to the original 100. mL of KI solution. Original amount of K + 100.0 mL KI 0.150 mol KI1 mol K + = .0150 mol K + 1000 mL 1 mol KI Volume Adjustment–new volume is 100 mL + 350 mL=450 mL 1000 ml 0.0150 mol K += 0.0333 M 450 mL

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