ACID AND ALKALIS計數
2017/07/03 22:18
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標題:
ACID AND ALKALIS計數
發問:
1. 2.63g of anhydrous sodium hydrogencarbonate are dissolved in water and the solution is made up to 250.0cm^3. 25.0cm^3 of this solution require 28.9 cm^3 of hydrochloric acid for complete reaction . What is the molarity of the acid?(H=1.0, C=12.0, O=16.0, Na=23.0)2. 0.200mol dm^-3 dilute hydrochloric acid... 顯示更多 1. 2.63g of anhydrous sodium hydrogencarbonate are dissolved in water and the solution is made up to 250.0cm^3. 25.0cm^3 of this solution require 28.9 cm^3 of hydrochloric acid for complete reaction . What is the molarity of the acid? (H=1.0, C=12.0, O=16.0, Na=23.0) 2. 0.200mol dm^-3 dilute hydrochloric acid was added to 25.0cm^3 of 0.400mol dm^-3 calcium hydroxide solution until complete neutralization. What is the concentration of calcium chloride in the resulting solution? 3. 100.0cm^3 of a solution contain 1.33g of a mixture of anhydrous sodium carbonate and soluiotn chloride. 25.0 cm^3 of the solution require 25.0cm^3 of 0.100 mol dm^3 hydrochloric acid for comlete reaction. What is the percentage by mass of sodium carbonate in the mixture? (C=12.0, O=16.0, Na=23.0) How to get the answers with: A1. 0.108 mol dm^-3 A2. 0.0800mol dm^-3 A3. 39.8%
最佳解答:
1. No. of moles of 2.63 g NaHCO3 = 2.63/(23.0 + 1.0 + 12.0 + 16x3) =0.0313 mol No. of moles of NaHCO3 used in titration = 0.0313 x (25/250) =0.00313 mol The titration : NaHCO3 + HCl → NaCl + H2O + CO2 No. of moles of NaHCO3 used = 0.00313 mol No. of moles of HCl used = 0.00313 mol Molarity of HCl = 0.00313 / (28.9/1000) = 0.108 M 2. 2HCl + Ca(OH)2 → CaCl2 + 2H2O No. of moles of HCl = 0.200 x (25.0/1000) = 0.005 mol No. of moles of Ca(OH)2 = 0.005 x (1/2) = 0.0025 mol Volume of Ca(OH)2 used = 0.0025 / 0.400 = 0.00625 dm3 No. of moles of CaCl2 formed = 0.005 x (1/2) = 0.0025 mol Volume of final solution = (25.0/1000) + 0.00625 = 0.03125 dm3 Concentration of CaCl2 = 0.0025 / 0.03125 = 0.0800 M 3. Na2CO3 + 2HCl → 2NaCl + H2O +CO2 No. of moles of HCl used = 0.100 x (25.0/1000) = 0.0025 mol No. of moles of Na2CO3 used in titration = 0.0025 x (1/2)= 0.00125 mol No. of moles of Na2CO3 in the mixture = 0.00125 x(100/25) = 0.005 mol Mass of Na2CO3 in the mixture = 0.005 x (23.0x2 + 12.0 +16.0x3) = 0.53 g % by mass of Na2CO3 in the mixture = (0.53/1.33) x 100% = 39.8%
其他解答:
ACID AND ALKALIS計數
發問:
1. 2.63g of anhydrous sodium hydrogencarbonate are dissolved in water and the solution is made up to 250.0cm^3. 25.0cm^3 of this solution require 28.9 cm^3 of hydrochloric acid for complete reaction . What is the molarity of the acid?(H=1.0, C=12.0, O=16.0, Na=23.0)2. 0.200mol dm^-3 dilute hydrochloric acid... 顯示更多 1. 2.63g of anhydrous sodium hydrogencarbonate are dissolved in water and the solution is made up to 250.0cm^3. 25.0cm^3 of this solution require 28.9 cm^3 of hydrochloric acid for complete reaction . What is the molarity of the acid? (H=1.0, C=12.0, O=16.0, Na=23.0) 2. 0.200mol dm^-3 dilute hydrochloric acid was added to 25.0cm^3 of 0.400mol dm^-3 calcium hydroxide solution until complete neutralization. What is the concentration of calcium chloride in the resulting solution? 3. 100.0cm^3 of a solution contain 1.33g of a mixture of anhydrous sodium carbonate and soluiotn chloride. 25.0 cm^3 of the solution require 25.0cm^3 of 0.100 mol dm^3 hydrochloric acid for comlete reaction. What is the percentage by mass of sodium carbonate in the mixture? (C=12.0, O=16.0, Na=23.0) How to get the answers with: A1. 0.108 mol dm^-3 A2. 0.0800mol dm^-3 A3. 39.8%
最佳解答:
1. No. of moles of 2.63 g NaHCO3 = 2.63/(23.0 + 1.0 + 12.0 + 16x3) =0.0313 mol No. of moles of NaHCO3 used in titration = 0.0313 x (25/250) =0.00313 mol The titration : NaHCO3 + HCl → NaCl + H2O + CO2 No. of moles of NaHCO3 used = 0.00313 mol No. of moles of HCl used = 0.00313 mol Molarity of HCl = 0.00313 / (28.9/1000) = 0.108 M 2. 2HCl + Ca(OH)2 → CaCl2 + 2H2O No. of moles of HCl = 0.200 x (25.0/1000) = 0.005 mol No. of moles of Ca(OH)2 = 0.005 x (1/2) = 0.0025 mol Volume of Ca(OH)2 used = 0.0025 / 0.400 = 0.00625 dm3 No. of moles of CaCl2 formed = 0.005 x (1/2) = 0.0025 mol Volume of final solution = (25.0/1000) + 0.00625 = 0.03125 dm3 Concentration of CaCl2 = 0.0025 / 0.03125 = 0.0800 M 3. Na2CO3 + 2HCl → 2NaCl + H2O +CO2 No. of moles of HCl used = 0.100 x (25.0/1000) = 0.0025 mol No. of moles of Na2CO3 used in titration = 0.0025 x (1/2)= 0.00125 mol No. of moles of Na2CO3 in the mixture = 0.00125 x(100/25) = 0.005 mol Mass of Na2CO3 in the mixture = 0.005 x (23.0x2 + 12.0 +16.0x3) = 0.53 g % by mass of Na2CO3 in the mixture = (0.53/1.33) x 100% = 39.8%
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