Question

a) An iron sulfide contains 36.5% S by mass. The iron sulfide is heated in an atmosphere of pure oxygen, which produces SO2(g) and an iron oxide containing 27.6% O by mass.

Write a balanced equation for the reaction.

If 1.0 kg of the iron sulfide reacts with excess oxygen, what is the theoretical yield of the iron oxide?

b) In the manufacture of Portland cement, limestone (CaCO3(s)) is decomposed into lime (CaO(s)) and CO2(g), in a kiln. Use data from Appendix C in your textbook to calculate how much heat is required to decompose 2.70 × 103 kg of limestone. (Assume that the heats of reaction are temperature independent.)

Answer 1

a)An iron sulfide contains 36.5% S by mass. The iron sulfide is heated in an atmosphere of pure oxygen, which produces SO2(g) and an iron oxide containing 27.6% O by mass.

Solution :- Iron sulfide that contains the 36.5 % S by mass is the FeS and the iron oxide that contains 27.6 % has formula Fe3O4

Write a balanced equation for the reaction.

3FeS + 5O2 ---- > Fe3O4 + 3SO2

If 1.0 kg of the iron sulfide reacts with excess oxygen, what is the theoretical yield of the iron oxide?

Now lets calculate the theoretical yield of the Fe3O4 using the mole ratio of the FeS and Fe3O4 in the balanced equation.

Lets first calculate moles of the 1.0 kg FeS

(1.0kg * 1000 g / 1 kg )* (1 mol FeS / 87.91 g) = 11.375 mol FeS

Now let’s calculate moles of Fe3O4 using the mole ratio

(11.375 mol FeS * 1 mol Fe3O4) / 3 mol FeS = 3.79 mol Fe3O4

Now lets convert moles of Fe3O4 to its mass

Mass = moles*molar mass

Molar mass of Fe3O4 = 231.5326 g/mol

Mass of Fe3O4 = 3.79 mol * 231.5326 g per mol

                           = 877.5 g Fe3O4

So the theoretical yield of the iron oxide = 877.5 g

b) In the manufacture of Portland cement, limestone (CaCO3(s)) is decomposed into lime (CaO(s)) and CO2(g), in a kiln. Use data from Appendix C in your textbook to calculate how much heat is required to decompose 2.70 × 103 kg of limestone. (Assume that the heats of reaction are temperature independent.)

Solution :-

Mass of CaCO3 = 2.70*10³ kg

Amount of heat needed to decomposition = ?

Lets first write the balanced reaction equation for the decomposition of the CaCO3

CaCO3(s)   ------ > CaO(s) + CO2(g)

Now lets calculate the enthalpy change for the reaction using the standard enthalpies of formation.

Formula

∆H^or = ∑∆H^of product - ∑∆H^of reactant

∆H^or = [(∆H^of CaO*1)+( ∆H^of CO2 *1) ] – [∆H^of CaCO3*1]

Let’s put the values in the formula

∆H^or = [(∆H^of CaO*1)+( ∆H^of CO2 *1) ] – [∆H^of CaCO3*1]

         = [(-635.1 kJ per mol *1)+(-393.5 kJ per mol *1)] – [-1206.9 kJ per mol *1]

         = 178.3 kJ/mol

So 1 mole of CaCO3(s) need 178.3 kJ heat

Now lets calculate for the 2.70*10³ kg CaCO3

(2.70*10³ kg * 1000 g / 1 kg )* (178.3 kJ / 100.0869 g ) = 4.81*10⁶ kJ

Therefore amount of heat needed for the decomposition of the 2.70*10³ kg CaCO3 =4.81*10⁶ kJ