Question

1. Many important metals occur as sulfide, arsenide, and antimonide minerals, especially in the Sudbury mineral complex. The first step in processing these ores involves “roasting” the ore in air to produce the metal or metal oxide, along with nonmetal oxides that can be serious pollutants if not trapped. Suppose that you roast 2.00 kg of the mineral marcasite, FeS2. The balanced equation for the reaction is: 4 FeS2 + 11 02 → 2 Fe2O3 + 8 SO2 (a) How many kg of Fe2O3 can be produced? (b) How many kg of the pollutant SO2 can be produced? (c) How many liters of air at 25°C are required for roasting? Assume that air contains 23% O2 by mass, and that the density of air at 25°C is 1.2 g/L. Molar masses: FeS2 120.0 g/mol Fe203 159.69 g/mol SO2 64.06 g/mol 2. A mixture of Na2SO4 and its hydrate Na2SO4.10H2O has a mass of 1.3598 g. After heating to drive off all the water, the mass is only 1.0057 g. What is the weight percentage of Na2SO4.10H20 in the original mixture? Report your answer to 4 significant figures. Molar masses: Na2SO4 142.04 g/mol; Na2SO4.10H20 322.19 g/mol. 3. FeCl3 reacts with AgNO3 in aqueous solutions to produce a precipitate of AgCl that can be filtered and weighed. The balanced equation for the reaction is: FeCl3 (aq) + 3 AgNO3 (aq) → 3 AgCl (s) + Fe(NO3)3 (aq) Suppose you have a mixture that contains FeCl3, plus other compounds that do not react with AgNO3. If 0.4158 g of the mixture yields 0.9489 g of AgCl, what is the percentage of FeCl3 in the mixture? Molar masses: FeCl3 162.21 g/mol AgCl 143.32 g/mol

Answer 1

1) a) The mass of Fe2O3 is calculated:

m Fe2O3 = 2 Kg FeS2 * (1 kmol / 120 kg) * (2 kmol Fe2O3 / 4 kmol FeS2) * (160 kg Fe2O3 / 1 kmol) = 1.33 kg

b) The mass of SO2 is calculated:

m SO2 = 2 Kg FeS2 * (1 kmol / 120 kg) * (8 kmol SO2 / 4 kmol FeS2) * (64 kg SO2 / 1 kmol) = 2.13 kg

c) The liters of air are calculated:

L Air = 2 Kg FeS2 * (1 kmol / 120 kg) * (11 kmol O2 / 4 kmol FeS2) * (32 kg O2 / 1 kmol) * (100 kg air / 23 kg O2) * (1000 g air / 1 kg) * (1 L air / 1.2 g) = 5314 L air

2) The mass of evaporated water is calculated:

m water = mi - mf = 1.3598 g - 1.0057 g = 0.3541 g

The hydrate mass is silenced:

m hydrate = 0.3541 g water * (322.19 g hydrate / 180 g water) = 0.6338 g

The percentage is calculated:

% hydrate = 0.6338 * 100 / 1.3598 = 46.61%

3) The mass of FeCl3 that forms the AgCl is calculated:

m FeCl3 = 0.9489 g AgCl * (1 mol AgCl / 143.32 g) * (1 mol FeCl3 / 3 mol AgCl) * (162.2 g FeCl3 / 1 mol) = 0.3580 g

The percentage of FeCl3 is calculated:

% FeCl3 = 0.3580 * 100 / 0.4158 = 86.10%

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