IN ORDER TO FIND OUT THE VALUE OF x AND y IN SAMPLE
(FORMULA=MgCl2 xKCl yK2SO4),A STUDENT CONDUCTED THE FOLLOWING EXPERIMENTS:
1) 10g SAMPLE WAS DISSOLVED IN 50cm^3 H2O , WHEN EXCESS SODIUM HYDROXIDE SOLUTION WAS ADDED TO THIS SOLUTION; 1.392g OF WHITE PRECIPITATE WAS OBTAINED
2) 5g SAMPLE WAS DISSOLVED IN 50cm^3 H2O, WHEN EXCESS CALCIUM BROMIDE SOLUTION WAS ADDED TO THIS SOLUTION; 1.626 g OF WHITE PRECIPITATE WAS OBTAINED.
CALCULATE THE VALUE OF x AND y
CHEM acid
2013-03-22 6:15 am
回答 (1)
2013-03-22 8:25 pm
✔ 最佳答案
Molar mass of MgCl2 = 24.3 + 35.5*2 = 95.3 g/molMolar mass of KCl = 39 + 35.5 = 74.5 g/mol
Molar mass of K2SO4 = 39*2 + 32.1 + 16*4 = 174.1 g/mol
Molar mass MgCl2•xKCl•yK2SO4 = (95.3 + 74.5x +174.1y) g/mol
Molar mass of Mg(OH)2 = 24.3 + (16 + 1)*2 = 58.3 g/mol
Molar mass of CaSO4 = 40 + 32.1 + 16*4 = 136.1 g/mol
Consider the reaction with excess NaOH(aq) :
MgCl2(aq) + 2NaOH(aq) → Mg(OH)2(s) + 2NaCl(aq)
No. of moles of Mg(OH)2 formed = 1.392/58.3 = 0.02388 mol
No. of moles of MgCl2 in the solution = 0.02388 mol
No. of MgCl2 in 10 g MgCl2•xKCl•yK2SO4= 0.02388 mol
Molar mass of MgCl2•xKCl•yK2SO4 = 10/0.02388 =418.7 g/mol
Consider the reaction with CaBr2(aq) :
K2SO4(aq) + CaBr2(aq) → CaSO4(s) +KBr(aq)
No. of moles of CaSO4 formed = 1.626/136.1 = 0.01195 mol
No. of moles of MgCl2•xKCl•yK2SO4 used =5/418.7 = 0.01194 mol
y = 0.01195/0.01194 = 1
Molar mass of MgCl2•xKCl•K2SO4 :
95.3 + 74.5x + 174.1 = 418.7
74.5x = 149.3
x = 2
Hence, x = 2 and y = 1
參考: Adam
收錄日期: 2021-04-16 15:48:35
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https://hk.answers.yahoo.com/question/index?qid=20130321000051KK00374