Part B Assuming that the coal was 1.90 % sulfur by mass and that combustion was

Question

Part B

Assuming that the coal was 1.90 % sulfur by mass and that combustion was complete, calculate the number of tons of sulfur dioxide produced by the plant during the year.

Express your answer numerically in tons.

Coal is a fossil fuel made primarily of carbon, but it also contains sulfur among other elements. When carbon, C, burns in air, it reacts with oxygen, O2, to produce carbon dioxide,CO2: C(s)+O2(g)CO2(g) When sulfur, S, burns in air, it reacts with oxygen, O2, to produce sulfur dioxide, SO2: S(s)+O2(g)SO2(g) In 1986 an electrical power plant in Taylorsville, Georgia, burned 8,376,726 tons of coal, a national record at that time.

Part A Assuming that the coal was 83.6 % carbon by mass and that combustion was complete, calculate the number of tons of carbon dioxide produced by the plant during the year. Express your answer numerically in tons.

Explanation / Answer

PART B:

Sulphure reacts as

S + O2 ---------> SO2

As per stiochiometry of the reaction one mole or 32g of S produce one mole of or 64g of SO2

Coal burnt in the year (1986 ) = 8,376,726 tons

Amount S in the coal = 1.9%

that means 100 tons of coal contain 1.9 tons of S

Amount of S in 8,376,726 tons of cola = (1.9 / 100) X 8,376,726 = 159157.8 tons of S

As 32g of S produce 64g of SO2 , It implies that 32 tons of S will produce 64 tons of SO2

Hence 1 ton of S will produce 64/32 = 2 tons of SO2

Therefore 159157.8 tons of S will produce 2 X 159157.8 f = 318,315.6 g of SO2

PART A:

This problem can be solved by following the same procedure

C + O2 ---------> CO2

1 mole or 12 g of C produce 1 mole or 44 g of CO2

12 tons of C produce 44 tons of CO2

Amount of C in the coal burnt in year 1986 = 8,376,72 X (83.6/100) = 700,293.79 tons of C.

700,293.79 tons of C will produce 700,293.79 X (44/12) = 2567743.9 tons of CO2.

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