The chemical formula of a metal carbonate is X2CO3. 1.59g of the carbonate are dissolved in 50cm^3 of distilled water. This solution requires 20cm^3 of 1.5M hydrochloric acid for complete reaction. What is the relative atomic mass of metal X?
Magnesium hydroxide is the active ingredient of an antacid. A 6g tablet of this antacid is dissolved in waater and made up to 250cm^3. 25cm^3 of the solution are titrated against 1.45M hydrochloric acid. 7cm^3 of the acid are required for complete neutralization. What is the percentage by mass of magnesium hydroxide in a tablet of the antacid?
8.3g of a hydrated sodium carbonate Na2CO3-nH2O are dissloved in water and the solution is made up to 250cm^3. 25cm^3 of this solution require 29cm^3 of 0.2M hydrochloric acid for complete reaction. What is the value of n?
25cm^3 of a 0.2M acid solution require 30cm^3 of 0.5M potassium hydroxide solution for complete neutralization. What is the basicity of the acid?
chemistry questions
2010-04-06 1:27 am
回答 (1)
2010-04-06 2:14 am
✔ 最佳答案
The chemical formula of a metal carbonate ......X2CO3 + 2HCl → 2XCl + H2O + CO2
No. of moles of HCl = 1.5 x (20/1000) = 0.03 mol
No. of moles of X2CO3 = 0.03 x (1/2) = 0.015 mol
Molar mass of X2CO3 = 1.59/0.015 = 106 g/mol
Let y be the relative atomic mass of X.
Molar mass of X2CO3:
2y + 12 + 16x3 = 106
y = 23
Relative atomic mass of X = 23
Magnesium hydroxide is ......
Mg(OH)2 + 2HCl → MgCl2 + H2O + CO2
No. of moles of HCl in titration = 1.45 x (7/1000) = 0.01015 mol
No of moles of Mg(OH)2 in titration = 0.01015 x (1/2) = 0.005075 mol
Mass of Mg(OH)2 in titration = 0.005075 x (24.3 + 16x2 + 1x2) = 0.2959 g
Mass of Mg(OH)2 in tablet = 0.2959 x (250/25) = 2.959 g
% by mass of Mg(OH)2 in tablet = (2.959/6) x 100% = 49.3%
8.3g of a hydrated sodium carbonate ......
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
No. of moles of HCl in titration = 0.2 x (29/1000) = 0.0058 mol
No. of moles of Na2CO3 in titration = 0.0058 x (1/2) = 0.0029 mol
Total no. of moles of Na2CO3•nH2O used = 0.0029 x (250/25) = 0.029 mol
Molar mass of Na2CO3•nH2O = 8.3/0.029 = 286
Molar mass of Na2CO3•nH2O:
23x2 + 12 + 16x4 + n(1x2 + 16) = 286
106 + 18n = 286
18n = 180
n = 10
25 cm³ of a 0.2 M acid solution ......
Let n be the bascity of the acid, and denote the acid as HnA.
HnA + nKOH → KnA + nH2O
No. of moles of KOH = 0.5 x (30/1000) = 0.015 mol
No. of moles of HnA = 0.2 x (25/1000) = 0.005 mol
Mole ratio
HnA : KOH = 1 : n
0.005 : 0.015 = 1 : n
1 : 3 = 1 : n
Basicity of the acid, n = 3
參考: 胡雪八
收錄日期: 2021-05-01 13:10:22
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