標題:
21st Century Chemistry 2 P.53
發問:
The stages involved in the Contact process can be represented by the following equations:Stage 1 S+O2----->SO2Stage 2 2SO2+O2----->2SO3(reversible)Stage 3 SO3+H2SO4----->H2S2O7Stage 4 H2S2O7+H2O----->2H2SO4If 8.00 kg of sulphur are used in stage 1,b) what is the mass of... 顯示更多 The stages involved in the Contact process can be represented by the following equations: Stage 1 S+O2----->SO2 Stage 2 2SO2+O2----->2SO3(reversible) Stage 3 SO3+H2SO4----->H2S2O7 Stage 4 H2S2O7+H2O----->2H2SO4 If 8.00 kg of sulphur are used in stage 1, b) what is the mass of sulphuric acid produced? I have some problems in the solution on P.54.It said that 'We can represent the whole process for the production of sulphuric acid by the following equation: S----->H2SO4(not a balanced chemical equation)' Why S----->H2SO4,not S----->2H2SO4?If I use S----->2H2SO4,will I calculate the wrong thing?
最佳解答:
In stage 3, H2SO4 is used. Therefore, the whole process can be represented as: S + H2SO4 → 2H2SO4 Or simply, S → H2SO4 If you use S → 2H2SO4, you will get a wrong answer. The calculations: Molar mass of S = 32.1 g mol-1 Molar mass of H2SO4 = 1x2 + 32.1 + 16x4 = 98.1 g mol-1 S → H2SO4 Mole ratio S : H2SO4 = 1 : 1 No. of moles of S used = 8000/32.1 = 249.2 mol No. of moles of H2SO4 formed = 249.2 mol Mass of H2SO4 formed = 249.2 x 98.1 = 24450 g = 24.45 kg =
其他解答:
點解係S----->....----->H2SO4, 而唔係S----->....----->2H2SO4 原因非常簡單 因為1mol的S只出到1mol的H2SO4 而你寫eqt當然係會見到S----->....----->2H2SO4 點解會咁? 因為你係尾2果個step加左conc sul落去嘛!!
21st Century Chemistry 2 P.53
發問:
The stages involved in the Contact process can be represented by the following equations:Stage 1 S+O2----->SO2Stage 2 2SO2+O2----->2SO3(reversible)Stage 3 SO3+H2SO4----->H2S2O7Stage 4 H2S2O7+H2O----->2H2SO4If 8.00 kg of sulphur are used in stage 1,b) what is the mass of... 顯示更多 The stages involved in the Contact process can be represented by the following equations: Stage 1 S+O2----->SO2 Stage 2 2SO2+O2----->2SO3(reversible) Stage 3 SO3+H2SO4----->H2S2O7 Stage 4 H2S2O7+H2O----->2H2SO4 If 8.00 kg of sulphur are used in stage 1, b) what is the mass of sulphuric acid produced? I have some problems in the solution on P.54.It said that 'We can represent the whole process for the production of sulphuric acid by the following equation: S----->H2SO4(not a balanced chemical equation)' Why S----->H2SO4,not S----->2H2SO4?If I use S----->2H2SO4,will I calculate the wrong thing?
最佳解答:
In stage 3, H2SO4 is used. Therefore, the whole process can be represented as: S + H2SO4 → 2H2SO4 Or simply, S → H2SO4 If you use S → 2H2SO4, you will get a wrong answer. The calculations: Molar mass of S = 32.1 g mol-1 Molar mass of H2SO4 = 1x2 + 32.1 + 16x4 = 98.1 g mol-1 S → H2SO4 Mole ratio S : H2SO4 = 1 : 1 No. of moles of S used = 8000/32.1 = 249.2 mol No. of moles of H2SO4 formed = 249.2 mol Mass of H2SO4 formed = 249.2 x 98.1 = 24450 g = 24.45 kg =
其他解答:
點解係S----->....----->H2SO4, 而唔係S----->....----->2H2SO4 原因非常簡單 因為1mol的S只出到1mol的H2SO4 而你寫eqt當然係會見到S----->....----->2H2SO4 點解會咁? 因為你係尾2果個step加左conc sul落去嘛!!
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