inr1 1/11/11/2 0/1 az ni in Points 0,1 11/15(73.3%) Assignment Submission this a

Question

inr1 1/11/11/2 0/1 az ni in Points 0,1 11/15(73.3%) Assignment Submission this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit c Assignment Scoring For this assignment, you submit answers by question parts. The nu Your best submission for each question part is used for your score. 1.1/2 points | Previous Answers Silber7 3.F.020A Solving Limiting-Reactant Problems for Reactions in Solution FOLLOW-UP PROBLEM 3.20 Even though gasoline sold in the United States no longer contains lead, this metal persists in the environment as a poison. Despite their toxicity, many compounds of lead are still used to make pigments. (a) What volume of 1.50 M lead(II) acetate contains 0.500 mol of Pb2+ ions? HINTS Getting Started I'm Stuck (b) When this volume reacts with 145 mL of 3.40 M sodium chloride, how many grams of solid lead chloride can form? (Sodium acetate solution also forms.) Supporting Materials l Appendices Constants and Periodic Table 25

Explanation / Answer

a) CH4(g) + 4Cl2(g) ----> CCl4(l) + 4HCl(g)

delta Hrxn = delta H0f CCl4 + 4*delta H0f HCl - 4*delta H0f Cl2 - delta H0f CH4 = -135.4 + 4*(-92.3) - (4*0) - (-74.9) = -429.7 kJ

b) 2H2S(g) + 3O2(g) ----> 2SO2(g) + 2H2O(g)

delta Hrxn = 2*delta H0f SO2 + 2*delta H0f H2O - 2*delta H0f H2S - 3*delta H0f O2 = 2*(-296.84) + 2*(-241.82) - 2*(-20.63) - 3*0 = -1036.06 kJ

c) N2(g) + 3H2(g) ----> 2NH3

delta Hrxn = 2*delta H0f NH3 - 3*delta H0f H2 - delta H0f N2 = 2*(-45.9) - 3*0 - 0 = -91.8 kJ

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.