c20_m7_l01

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Chemistry

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May 4, 2024

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D5 MODULE 7: LESSON 1 ASSIGNMENT CHEM 20 (?/34 MARKS) Lesson 1 Assignment: Limiting Reagents Lab: Chemical Analysis Using a Graph Read “Lab Exercise 8.A: Chemical Analysis Using a Graph” on page 317 in your textbook. You may also find the information about this reaction on page 318 useful in completing your analysis. After you have read the lab, answer the following questions to complete your analysis. Analysis (2 marks) 1. Write the balanced chemical equation for the reaction described in this lab exercise. Answer: Pb(NO 3 ) 2 (aq) + 2KI(aq) -> PbI 2 (s) + 2KNO 3 (aq) (1 mark) 2. Identify the limiting reagent and the excess reagent in this lab. Answer: Limiting Reagent = potassium iodide Excess Reagent = lead (II) nitrate (2 marks) 3. What is observed when the two reacting solutions are mixed? Describe a test that could be performed to ensure that the reactions performed were complete. Answer: - When the two reacting solutions, lead (II) nitrate and potassium iodide are mixed then a bright yellow precipitate is formed. To ensure that the reactions performed were complete them add more of the limiting reagent (potassium iodide) using a medicine dropper into the solution. If any there is still cloudiness that means that the reaction isn’t yet complete. So, continue to add more drops of the limiting reagent (potassium iodide) until no more precipitate forms, because at that point the reaction is complete. 4. Use a spreadsheet to create a graph of the data shown in Table 1. In this lab, the manipulated variable is the mass of Pb(NO 3 ) 2 reacting. Use your completed graph to answer questions 4.a. and 4.b. (1 mark) a. A lab technician weighs out 2.5 g of Pb(NO 3 ) 2 (s) and dissolves it in water. What mass of PbI 2 (s) can be produced from this solution? Answer: Pb(NO 3 ) 2 (aq) + 2KI(aq) → 2KNO 3 (aq) + PbI 2 (s) 2.5 g. m ? 331.2 g/mol 461.01 g/mol n[Pb(NO 3 ) 2 ] = (2.5 g) / (331.2 g/mol) n= 0.00754830917 mol
n(PbI 2 )= (0.00754830917 mol)(1) n= 0.00754830917 mol m(PbI 2 )= (0.00754830917 mol)(461.01 g/mol) m= 3.48 g of PbI 2 can be produced (1 mark) b. Explain how the data shown on the graph could be used to identify the limiting reagent in the reaction being studied. Answer: - When you graph the data, the amount of potassium iodide produced increases, and so does the mass of iron (ii) nitrate until no more lead iodide is produced, because all of the potassium iodide is consumed in the reaction. This would mean that all the potassium iodide has completely reacted and it would be the limiting reagent. Assignment Questions 1. 25.0 mL of 0.150-mol/L KBr(aq) is mixed with excess Pb(NO 3 ) 2 (aq). (2 marks) a. Write the balanced chemical equation for this reaction. Answer: 2KBr(aq) + Pb(NO 3 ) 2 (aq) -> 2KNO 3 (aq) + PbBr 2 (s) (2 marks) b. Calculate the number of moles of the limiting reagent. Answer: n(KBr) = 0.0250L × 0.150 mol/L = 0.00375mol (1 mark) c. Predict the number of moles of precipitate. Answer: N(PbBr 2 ) = 0.00375/2 = 0.001875 mol (2 marks) d. Calculate the mass of precipitate. Answer: m(PbBr 2 )= (0.001875 mol)(367.01g/mol) = 0.690g 2. 3.00 L of methane, CH 4 (g), is combusted with 5.50 L of oxygen gas. Assume SATP conditions. (2 marks) a. Write the balanced chemical equation. Answer: CH 4 (g) + 2O 2 (g) -> CO 2 (g) + 2H 2 O(l) (2 marks) b. Determine the number of moles of each reactant. Answer: CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) 3.00 L 5.50 L
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