A power plant burns 100 metric tons of coal per day with a sulfur content of 5%. The coal contains 3% ash (noncombustible portion). Samples of the ash show that it has a sulfur content of 10 % sulfur by weight.
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Answer:
Mass of #CaSO_3# produced daily#=##1.70xx10^8g#
Explanation:
#SO_2(g) + CaO(s) rarr CaSO_3(s)#
#"1 US ton"# #=# #907.19*kg#
#"Mass of coal(kg)"# #=##2000*"ton"xx907*kg*"ton"^-1# #=##1.814xx10^6*kg#.
#"Mass of sulfur"# #=##1.814xx10^6*kgxx2.5%# #=# #45,350*kg#
#"Moles of sulfur"# #=# #(45,350*kgxx10^3*g*kg^-1)/(32.06*g*mol^-1)##=# #1.41xx10^6*mol#.
And thus we may have #1.41xx10^6*mol##SO_2# produced during the daily operation.
Given the equivalence of sulfur dioxide and calcium oxide in the first equation, clearly we need #1.41xx10^6*mol# #CaO# to produce #1.41xx10^6*mol# #CaSO_3#.
#"Mass of calcium sulfite produced daily"# #=##1.41xx10^6*molxx120.17*g*mol^-1# #=##1.70xx10^8*g# #=# #1.70xx10^5*kg#.
There is a lot of arithmetic here; please go over my calculations; there is no money back guarantee.
Mass of #CaSO_3# produced daily#=##1.70xx10^8g#
Explanation:
#SO_2(g) + CaO(s) rarr CaSO_3(s)#
#"1 US ton"# #=# #907.19*kg#
#"Mass of coal(kg)"# #=##2000*"ton"xx907*kg*"ton"^-1# #=##1.814xx10^6*kg#.
#"Mass of sulfur"# #=##1.814xx10^6*kgxx2.5%# #=# #45,350*kg#
#"Moles of sulfur"# #=# #(45,350*kgxx10^3*g*kg^-1)/(32.06*g*mol^-1)##=# #1.41xx10^6*mol#.
And thus we may have #1.41xx10^6*mol##SO_2# produced during the daily operation.
Given the equivalence of sulfur dioxide and calcium oxide in the first equation, clearly we need #1.41xx10^6*mol# #CaO# to produce #1.41xx10^6*mol# #CaSO_3#.
#"Mass of calcium sulfite produced daily"# #=##1.41xx10^6*molxx120.17*g*mol^-1# #=##1.70xx10^8*g# #=# #1.70xx10^5*kg#.
There is a lot of arithmetic here; please go over my calculations; there is no money back guarantee.
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