please answer c and f ii) State four uses of caustic soda. iii) State the advant

Question

please answer c and f

ii) State four uses of caustic soda. iii) State the advantages of producing chlorine by electrolysis to that of Leblanc process. c) Calculate the standard enthalpy change per mole of Na_2CO_3 produced by the Leblanc process, using data below. From your calculation, what do you conclude about the overall heat requirements of that process? NaCL = 411.15 kjmol^-1 Na_2CO_3 = 1130.68 kjmol^-1 CH_2SO_4 = 0.850.67 kjmol^-1 CaCo_3 = 1206.92 kjmol^-1 CO_2(g) = 393.51 kjmol^-1 HCL = 92.31 kjmol^-1 C = 0 kjmol^-1 d. With the aid of flow diagram, describe the production of caustic soda by Leblanc process. b. With a good flow diagram describe the lead-chamber process of producing sulfuric (sulphuric) acid. f) A sample of Portland cement contains the following percentages by mass of oxides: CaO, 64.3%: SiO_2%, 21.21.: Al_2O_3%5.9%, fe_2O_3, 2.9%: MgO, 2.5%: SO_3, 1.8%: N_2O, 1.4%. Calculate the chemical amount of calcium, silicon, aluminum, iron, magnesium, sulfur and sodium atoms per mole of oxygen atoms in this sample.

Explanation / Answer

For letter c:

The leblanc process consists of the following three reactions:

2NaCl + H2SO4 -> Na2SO4 + 2HCl

Na2SO4 + 2C -> Na2S + 2CO2

Na2S + CaCO3 -> Na2CO3 + CaS

For which we obtain the individual heats of reaction:

deltaHro = Heats of formationproducts - Heats of formationreactants

For first reaction:

2NaCl + H2SO4 -> Na2SO4 + 2HCl

deltaHro = (-1380 kJ/mol + (2*-92.3 kJ/mol)) - (2*-411 kJ/mol + -814 kJ/mol)

deltaHro = -1564.6 kJ/mol + 1636 kJ/mol = 71.4 kJ/ mol

For second reaction:

Na2SO4 + 2C -> Na2S + 2CO2

deltaHro = (-437 kJ/mol + (2*-393.5 kJ/mol)) - (-1380 kJ/mol)

deltaHro = -1224 kJ/mol + 1380 kJ/mol = 156 kJ/ mol

For third reaction:

Na2S + CaCO3 -> Na2CO3 + CaS

deltaHro = (-1130.7 kJ/mol + (-473.21 kJ/mol)) - (-437 kJ/mol + -1206.9 kj/mol)

deltaHro = -1603.91 kJ/mol + 1643.9 kJ/mol = 40 kJ/ mol

Adding them up:

deltaHro = 267.4 kJ/mol

For f:

We consider a 100 gram sample of cement, from which we make the following table:

Total moles of Oxygen = 3.83392 moles

We now get percentage of atoms pero moles of oxygen:

Ca = 1.1468 moles Ca / 3.83392 moles O = 0.3

Si = 0.353 moles Si / 3.83392 moles O = 0.092

Al = 1.1574 moles Al / 3.83392 moles O = 0.3019

Fe = 0.03632 moles Fe / 3.83392 moles O = 0.0094

Mg = 0.0619 moles Fe / 3.83392 moles O = 0.016

S = 0.02248 moles S / 3.83392 moles O = 0.0059

N = 0.0608 moles N / 3.83392 moles O = 0.016

Molecule Percentage Mass Moles Moles of O2 CaO 64.3% 64.3 g 64.3 g * (1 mol / 56.07 g) = 1.1468 moles 1.1468 moles CaO * (1 mol O / 1 mol CaO) = 1.1468 moles SiO2 21.21% 21.21 g 21.21 g * (1 mol / 60.08 g) = 0.353 moles 0.353 moles SiO2 * (2 mol O / 1 mol SiO2) = 0.706 moles Al2O3 5.9% 5.9 g 5.9 g * (1 mol / 101.96 g) = 0.5787 moles 0.5787 moles Al2O3 * (3 mol O / 1 mol Al2O3) = 1.7361 moles Fe2O3 2.9% 2.9 g 2.9 g * (1 mol / 159.69 g) = 0.01816 moles 0.01816 moles Fe2O3 * (3 mol O / 1 mol Fe2O3) = 0.05448 moles MgO 2.5% 2.5 g 2.5 g * (1 mol / 40.34 g) = 0.0619 moles 0.0619 moles MgO * (1 mol O / 1 mol MgO) = 0.0619 moles SO3 1.8% 1.8 g 1.8 g * (1 mol / 80.06 g) = 0.02248 moles 0.02248 moles SO3 * (3 mol O / 1 mol SO3) = 0.06744 moles NO2 1.4% 1.4 g 1.4 g * (1 mol / 46.05 g) = 0.0304 moles 0.0304 moles NO2 * (2 mol O / 1 mol NO2) = 0.0608 moles

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

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