ASsighment Submission For this assignment, you submit answers by question parts.

Question

ASsighment Submission For this assignment, you submit answers by question parts. The number of submissions remaining for each question Assignment Scoring Your best submission for each question part is used for your score. 1. 1/2 points I Previous Answers Silber7 3.F.020A. Solving Limiting-Reactant Problems for Re 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? .333 (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.) 24

Explanation / Answer

Pb(C2H3O3)2 -------> Pb+2 + C2H3O2-2

1 mole of lead acetate contains 1 mole of Pb+2

0.5 moles of Pb+2 contains 0.5 moles of lead acetate

Molarity = moles/L

1.5 moles are there in 1L

0.5 moles correspond to 0.5/1.5= 0.333L

2.

Ca(s) +1/2O2(g) ------> CaO(s)    delH= -635.1 Kj   (1)

CaCO3------>CaO(s)+ CO2(g)   delH=178.3 Kj    (2)

Subtracting Eq.1 and Eq.2 gives

Ca+1/2O2+CO2(g) ------> CaCO3(g) delH=-635.1-178.3 = -813.4 KJ

3.

C(diamond)+O2(g)---------> CO2(g)   delH= -395.4 Kj (1)

2CO2(g) -----> 2CO(g) +O2(g)   delH= 566.0 Kj (2)

C(graphaite) +O2--------> CO2 delG= -393.5 KJ (3)

2CO(g)---> C(graphite)+ CO2(g)   delH= -172.5 Kj   (4)

Additing Eq.2 and 4 gives

CO2(g)--->C (graphite)+O2(g)   delH= 566-172.5= 393.5 (5)    ( reversed reactions of Eq.3)

Subteaction of Eq.1 and 3 gives

C(diamond)------>C(graphite)   delH= -395.4+393.5=-1.9 Kj

4.

For the reaction

C2H2+5/2O2----------> 2CO2(g) +H2O(g) delH= -1255.8 Kj

Enthalpy change fo the reaction = sum of enthalpies of products- sum of enthalpies of reactants

=2* enthalp of CO2+enthalpy of water-{ 1*Enthalpy of C2H2 + 2.5* enthalpy of oxygen}

Since enthalpy of oxygen=0

=-1255.8= 2*(-393.5) +(-241.8)- enthalpy of C2H2

Enthalpy of C2H2= -1028.8+1255.8=227 Kj/mol

5.

Balanced reaction is CH4(g) +4Cl2(g) --------à CCl4 (l) + 4HCl(g)

CH4(g) --à C(s) + 2H2(g) delta H = +74.6 kJ    (1)

C(s) + 2Cl2(g) à CCl4(g) delta H = -95.7kj   (2)

H2(g) + Cl2(g) à 2HCl(g) delta H = -184.6kj (3)

Multiply Eq.3 with 2

2H2(g) +2Cl2(g) -------à 4HCl (g) delH= -2*186.4=-372.8 Kj (3A)

Addition of Eq.1, Eq.2 and Eq.3 A gives

CH4+4Cl2-------> CCl4+4HCl (g) delH= 74.6-95.7-372.8=-393.9 Kj

Balanced reaction is CH4(g) +4Cl2(g) --------à CCl4 (l) + 4HCl(g)

CH4(g) --à C(s) + 2H2(g) delta H = +74.6 kJ    (1)

C(s) + 2Cl2(g) à CCl4(g) delta H = -95.7kj   (2)

H2(g) + Cl2(g) à 2HCl(g) delta H = -184.6kj (3)

Multiply Eq.3 with 2

2H2(g) +2Cl2(g) -------à 4HCl (g) delH= -2*186.4=-372.8 Kj (3A)

Addition of Eq.1, Eq.2 and Eq.3 A gives

CH4+4Cl2-------à CCl4+4HCl (g) delH= 74.6-95.7-372.8=-393.9 Kj

b) balanced reaction is 2H2S(g) +3O2(g) -------à 2SO2+2H2O

Enthalpy change of reaction= sum of enthalpies of products – sum of enthalpies of reactants

=2*enthalpy of H2sO(g) + 2* enthalpy of SO2- { 2* enthalpy of H2S +3* enthalpy of oxygen}

Enthalpy of oxygen =0 hence

=2*(-241.8)+2*(-296.8)- {2*(-20.630+3*0} =-1035.94 KJ

5.

Bond energy of Ntriple bond N= 942 Kj/mol and bond energy of H-H=432 Kj/mol and that of N-H =386 Kj/mol

Enthalpy change of reaction= Energy required to break the reactants- energy required to form products

= 942+3*432-(2*3*386)=-78 Kj/mol

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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