:
CHEM acid
:
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 OBTAINED2) 5g SAMPLE WAS DISSOLVED IN... 示更多 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
最佳解答:
Molar mass of MgCl2 = 24.3 + 35.5*2 = 95.3 g/mol Molar 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 MgCl-2 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
其他解答:
此文章自奇摩知+如有不便留言告知
< async src="/pagead2.googlesyndication.com/pagead/js/adsbygoogle.js">
13413D6BADCE0D57
文章定位：
