(1)0.12g of a divalent metal was dissolved by adding to it 30cm³of 0.25M HCl. The resulting solution required 27cm³ of 0.141M NaOH for complete neutralization. Determine the relative atomic mass of the metal.
(2)A mixture of KOH and K₂CO₃were dissolved in water and the solution made up to 250 cm³. 25 cm³ of this solution required 35.5 cm³ of 0.075M H₂SO₄ to neutralize it. 0.055g of CO₂ were collected. Find the composition by mass of the original mixture.
化學mole 計算
2008-06-15 7:58 am
回答 (1)
2008-06-15 9:11 am
✔ 最佳答案
(1)Consider the neutralization between the unreacted HCl and NaOH.
HCl + NaOH → NaCl + H2O
mole ratio HCl : NaOH = 1 : 1
No. of moles of NaOH used = MV = 0.141 x (27/1000) = 0.003807 mol
No. of moles of HCl used = 0.003807 mol
Total number of moles of HCl = MV = 0.25 x (30/1000) = 0.0075 mol
Consider the reaction between the metal (M) with HCl.
M + 2HCl → MCl2 + H2
Mole ratio M : HCl = 1 : 2
No. of moles of HCl used = 0.0075 - 0.003807 = 0.003693 mol
No. of moles of M used = 0.003693 x (1/2) = 0.001847 mol
Molar mass of M = mass/mol = 0.12/0.001847 = 65.0 g mol-1
Relative atomic mass of the metal (M) = 65.0
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(2)
Let:
No. of moles of KOH in the mixture = 10a mol
No. of moles of K2CO3 in the mixture = 10b mol
Out of the 250 cm3 of the solution, only 25 cm3 is used in titration. Therefore,
No. of moles of KOH in 25 cm3 of the solution = 10a x (25/250) = a mol
No. of moles of K2CO3 in 25 cm3 of the solution = 10b x (25/250) = b mol
Consider the neutralization between KOH and H2SO4:
2KOH + H2SO4 → K2SO4 + 2H2O
Mole ratio KOH : H2SO4 = 2 : 1
No. of moles of KOH used = a mol
No. of moles of H2SO4 used in this reaction = a x (1/2) = (a/2) mol
Consider the neutralization between K2CO3 and H2SO4.
K2CO3 + H2SO4 → K2SO4 + H2O + CO2
Mole ratio K2CO3 : H2SO4 : CO2 = 1 : 1 : 1
No. of moles of K2CO3 used = b mol
No. of moles of H2SO4 used in this reaction = b mol
No. of moles of CO2 formed = b mol
Molar mass of CO2 = 12 + 16x2 = 44 g mol-1
Consider the number of moles of CO2 formed:
b = 0.055/44
Hence, b = 0.00125
Consider the total number of moles of H2SO4 used:
(a/2) + b = 0.075 x (35.5/1000)
(a/2) + 0.00125 = 0.002663
Hence, a = 0.002826
No. of moles of KOH in the mixture = 10a = 0.02826 mol
Molar mass of KOH = 39 + 16 + 1 = 56 g mol-1
Mass of KOH in the mixture = 0.02826 x 56 = 1.583 g
No. of moles of K2CO3 in the mixture = 10b = 0.0125 mol
Molar mass of K2CO3 = 39x2 + 12 + 16x3 = 138 g mol-1
Mass of K2CO3 in the mixture = 0.0125 x 138 = 1.725 g
The mixture contains 1.583 g of KOH and 1.725 g of K2CO3.
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