Acid rain is generated when sulfur oxide compounds like SO_2 and SO_3 react with
ID: 478860 • Letter: A
Question
Acid rain is generated when sulfur oxide compounds like SO_2 and SO_3 react with water. Although some of sulfur oxide compounds in the environment come from natural sources. roughly 80% of the sulfur oxides present in the environment are a result of human activity, notably, the combustion of coal that contains sulfur. The annual emissions of sulfur dioxide from burning coal in the United States are approximately 26 million metric tons. (1 metric ton = 1000 kg) Use dimensional analysis and percent composition to solve these problems. It is possible to solve the problems using the concept of a mole, but do NOT use moles as part of your solutions. Determine how many grams of sulfur must be in the coal in the US to produce 26 million metric tons of SO_2. How many metric tons of coal must be burned in the US annually if 26 million metric tons of SO_2 is formed? Assume that the coal is 29% sulfur by mass. The US actually burns approximately 860 million metric tons of coal per year. Why is your value in part (b) different from this amount? Explain. How many pounds of SO_2 are released into the atmosphere during the combustion of enough coal to fuel a 100 W lightbulb for a year? It takes approximately 715 lbs. of coal to fuel a 100 W lightbulb for a year. Assume that the coal burned is 2% sulfur by mass and that all of the sulfur in the coal reacts to form SO_2. Complete these questions using percent com and dimensional analysis. Submit your responses in Canvas before the deadline at 11:55 pm on Sunday, February 5, 2017. The link for the submission is found under the information for Friday of this week. Your electronic submission must be a .pdf., .doc, .jpg or a .png file. Late submissions are not accepted.Explanation / Answer
Ans. Balanced reaction: S + O2 --------> SO2
Given, Mass of SO2 produced = 26 million metric tons
= 26 x 106 x 103 kg ; [1 million = 106, 1 metric ton = 103 kg]
= 2.6 x 1010 x 103 g ; [1 kg = 103 g]
= 2.6 x 1013 g
Number of moles of SO2 = Mass / molar mass of SO2
= 2.6 x 1013 g / (64.0648 g mol-1)
= 4.06 x 1011 mol
According to stoichiometry (see, top) of combustion of Sulfur, 1 mol S produces 1 mol SO2.
That is,
Number of moles of S = number of moles of SO2
= 4.06 x 1011 mol
Mass of S = number of moles of S x atomic mass of S
= 4.06 x 1011 mol x (32.066 g mol-1)
= 1.30 x 1013 g
Therefore, the amount of S present in coal = 1.30 x 1013 g
Ans. B. S + O2 --------> SO2
Atomic Mass of S = 32.066 g mol-1
Molar mass of SO2 = 64.0648 g mol-1
Since 1 mol S produces 1 mol SO2, hence-
64.0648 g is equivalent to 32.066 g S
Or, 1 g - - (32.066 / 64.0648) g
Or, 2.6 x 1013 g - - (32.066 / 64.0648) x 2.6 x 1013 g
= 1.30 x 1013 g
= 1.300 million metric ton
Let, the total mass of coal be X million metric tons.
Mass of S = 2% of X = 0.02X
Now,
0.02X = 1.300 million metric ton
Or, X = (1.300 / 0.02) million metric ton
= 65 million metric ton
= 6.5 x x107 metric tons
Hence, amount of coal = 6.5 x x107 metric tons
Ans. C. The values in part B (65 million metric tons of coal) is very less than the actual amount (860 million metric tons) because the industries using coal as fuel source have installed facilities (SO2 scrubbers) to minimize the released SO2 produced during combustion of coal.
Ans. D. Mass of S in 715 lb coal = 2% of 715 lb
= 14.3 lb
From stoichiometry of reaction-
32.066 g S is equivalent to 64.0648 g SO2
Or, 32.066 lb is equivalent to 64.0648 lb SO2 ; [changing the unit of mass does not affect the relative proportion of two equivalent weights].
So,
32.066 g lb is equivalent to 64.0648 lb SO2
Or, 1 lb - - (64.0648/ 32.066) lb SO2
Or, 14.3 lb - - (64.0648/ 32.066) x 14.3 lb SO2
= 28.57 lb
Therefore, amount of SO2 released during specified process = 28.57 lb
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