標題:
mole 問題
發問:
A) The stages involved in the Contact Process can be represented by the following equations:S(s) +O? (g)-->SO? (g)2SO? (g) +O?-->2SO? (g)SO? (g) +H?SO? (l)-->H?S?O7(l)H?S?O7(l) +H?O(l)-->2H?SO?Suppose that the conversion of sulphur to sulphuric acid is 100%.1) How many moles of sulphur are... 顯示更多 A) The stages involved in the Contact Process can be represented by the following equations: S(s) +O? (g)-->SO? (g) 2SO? (g) +O?-->2SO? (g) SO? (g) +H?SO? (l)-->H?S?O7(l) H?S?O7(l) +H?O(l)-->2H?SO? Suppose that the conversion of sulphur to sulphuric acid is 100%. 1) How many moles of sulphur are required to produce one mole of sulphuric acid? 2)If the annual production of sulphuric acid in this country is 2.6x10^9kg, calculate the annual consumption of sulphur, in kg, in the Contact Process. B) The relative molecular mass of an alkanol X (CnH?n+?OH) is 60.0. X contains 60% of carbon by mass. Calculate the number of moles of carbon in one mole of X and hence deduce the molecular formula of X.
最佳解答:
(A)(1) 1 mol of S is needed to produce 1 mol of H2SO4. Reason : Step 1 : 1 mol S → 1 mol SO2 Step 2 : 1 mol SO2 → 1 mol SO3 Step 3 : 1 mol SO3 + 1 mol H2SO4 → 1 mol H2S2O-7 Step 4 : 1 mol H2S2O7 → 2 mol H2SO4 Although there are 2 mol of H-2-SO4 produced in step 4, only 1 mol of H2SO4 is produced from S. This is because 1 mol of H-2SO4 is added in Step 3. (Both (A)(1) and (A)(2) are past HKCE questions.) ========== (A)(2) Molar mass of S = 32.1 g mol-1 Molar mass of H2SO4 = 2x1 + 32.1 + 16x4 = 98.1 g mol-1 Mole ratio of S : H2SO4 = 1 : 1 Mass of H2SO4 produced = 2.6 x 109 kg = 2.6 x 1012 mol No. of mol of H2SO4 produced = (2.6 x 1012) / 98.1 = 2.65 x 1010 mol Mass of S needed = (2.65 x 1010) x 32.1 = 8.51 x 1011 g = 8.51 x 108 kg ========== (B) In 1 mol of X (CnH2n+1OH) : Mass of X = Molar mass of X = 60 g Mass of C in X = Mass of n mol of C = 12n g (12n/60) x 100% = 60% Hence, n = 3 Molecular formula of X = CnH2n+1OH = C3H7OH
其他解答:
mole 問題
發問:
A) The stages involved in the Contact Process can be represented by the following equations:S(s) +O? (g)-->SO? (g)2SO? (g) +O?-->2SO? (g)SO? (g) +H?SO? (l)-->H?S?O7(l)H?S?O7(l) +H?O(l)-->2H?SO?Suppose that the conversion of sulphur to sulphuric acid is 100%.1) How many moles of sulphur are... 顯示更多 A) The stages involved in the Contact Process can be represented by the following equations: S(s) +O? (g)-->SO? (g) 2SO? (g) +O?-->2SO? (g) SO? (g) +H?SO? (l)-->H?S?O7(l) H?S?O7(l) +H?O(l)-->2H?SO? Suppose that the conversion of sulphur to sulphuric acid is 100%. 1) How many moles of sulphur are required to produce one mole of sulphuric acid? 2)If the annual production of sulphuric acid in this country is 2.6x10^9kg, calculate the annual consumption of sulphur, in kg, in the Contact Process. B) The relative molecular mass of an alkanol X (CnH?n+?OH) is 60.0. X contains 60% of carbon by mass. Calculate the number of moles of carbon in one mole of X and hence deduce the molecular formula of X.
最佳解答:
(A)(1) 1 mol of S is needed to produce 1 mol of H2SO4. Reason : Step 1 : 1 mol S → 1 mol SO2 Step 2 : 1 mol SO2 → 1 mol SO3 Step 3 : 1 mol SO3 + 1 mol H2SO4 → 1 mol H2S2O-7 Step 4 : 1 mol H2S2O7 → 2 mol H2SO4 Although there are 2 mol of H-2-SO4 produced in step 4, only 1 mol of H2SO4 is produced from S. This is because 1 mol of H-2SO4 is added in Step 3. (Both (A)(1) and (A)(2) are past HKCE questions.) ========== (A)(2) Molar mass of S = 32.1 g mol-1 Molar mass of H2SO4 = 2x1 + 32.1 + 16x4 = 98.1 g mol-1 Mole ratio of S : H2SO4 = 1 : 1 Mass of H2SO4 produced = 2.6 x 109 kg = 2.6 x 1012 mol No. of mol of H2SO4 produced = (2.6 x 1012) / 98.1 = 2.65 x 1010 mol Mass of S needed = (2.65 x 1010) x 32.1 = 8.51 x 1011 g = 8.51 x 108 kg ========== (B) In 1 mol of X (CnH2n+1OH) : Mass of X = Molar mass of X = 60 g Mass of C in X = Mass of n mol of C = 12n g (12n/60) x 100% = 60% Hence, n = 3 Molecular formula of X = CnH2n+1OH = C3H7OH
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A 1. 0.5 mole of sulphur is needed. 2. the number of mole of sulphuric acid =2.6x10^9 /98=2.65x10^7mole since the number of mole of sulphur is half of sulphuric acid, the no. of mole of sulphur is 2.65x10^7 x0.5 = 1.3x10^7 mole the mass of sulphur =1.3x10^7 x 32= 4.16x10^8 kg B the mass of carbon =60x 60% =36 the no of mole of carbon=36/12=3 the molecular formula is C3H7OH5AEEC82B53E1B405
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