For questions 20-22 refer to the production of dioxide (MM-6 sulfide 239.3 g/mol

Question

For questions 20-22 refer to the production of dioxide (MM-6 sulfide 239.3 g/mol) to commonly produced chemical by used produce 6Aog/moi) tom d to the balanced the chemical sulfuric chemica sulfur dioxide most 2Pbs(s) is according 20)The worldwide che each year. A is a metric unit typically produces imm equal to 1000 tonne ustry needed to produce 200 kilograms. How much lead sulfide is a, 200 million tonnes of sulfur dioxide if the percent yield is 100%? tonnes 200 x 10 kilograms ce )200 64.0 239.3 million tonnes 200 239.3 640 million tonnes 21)How many moles of oxygen are to fully react with 100 kg of sulfide? a. 0.6268 moles lead 6268 moles 22)How many kilograms of sulfur dioxide are made for every 100 kg of lead sulfide that reacts if the yield is 68%? 39 kg d. 108 kg 23)Based on the phase diagram shown below, which ofthe following statements are correc? int in the transformation that falls along a straight line L Sublimation occurs from point A to poi ere the gas and liquid phases are in equilibrium ll. Can E repres have a greater average kinetic energy than those at point F. Molecules at At point E, all three a. V b. I, III, IV l, ll, Ill P (atm) l, IV, V e II, IV GenCheml Spr 2017 Exam 3 Version A

Explanation / Answer

The balanced chemical equation is

2 PbS (s) + 3 O2 (g) ------> 2 PbO (s) + 2 SO2 (g)

20) As per the balanced stoichiometric equation,

2 mole PbS = 2 mole SO2

===> 1 mole PbS = 1 mole SO2

Molar mass of PbS = (1*207.2 + 1*32.065) g/mol = 239.265 g/mol = (239.265 g/mol)*(1 kg/1000 g) = 0.239265 kg/mol.

Molar mass of SO2 = (1*32.065 + 2*15.9994) g/mol = 64.0638 g/mol = (64.0638 g/mol)*(1 kg/1000 g) = 0.0640638 kg/mol.

We wish to have 200 million tonnes of SO2 = 200*106 tonnes = (200*106 tonnes)*(1000 kg/1 tonne) = 2.0*1011 kg.

Moles of SO2 produced = (2*1011 kg)*(1 mole/0.0640638 kg) = (2*1011 kg)*(1 mole/239.265 kg)*1000 = (2*1014/64.0638) mole (2*1014/64.0) mole

Moles of PbS required = (2*1014/64.0) mole SO2*(0.239265 kg/1 mole) PbS*(1 mole PbS/1 mole SO2) = (2*1011*239.265/64.0) kg (2*1011*239.3/64.0) kg = (2.0*1011*239.3/64.0)kg*(1 million/106)*(1 tonne/103 kg) = (200*239.3/64.0) million tonnes (ans).

Ans: (d)

21) As per the above stoichiometric reaction,

2 moles PbS = 3 moles O2

100 kg PbS = (100 kg PbS)*(1 mole/0.2393 kg) = (1*105/239.3) mole PbS

Moles of O2 required = (1*105/239.3) mole PbS*(3 moles O2/2 mole PbS) = 626.828 mole O2 = 686.8 moles.

Ans: (c)

22) The conversion ratio is 68%.

We have already deduced that 100 kg PbS = (1*105/239.3) mole PbS

Mass of SO2 generated = 68%*(1*105/239.3) mole PbS*(1 mole SO2/1 mole PbS)*(0.0640 kg SO2/1 mole) = (68/100)*(1*105*0.0640/239.3) kg = 18.186 kg 18 kg.

Ans: (a)

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