Đáp án:
Giải thích các bước giải: cau 4
nCl2 = 11,536/22,4 =0,515 mol
a, PTHH : 2KMnO4 + 16HCl -> 2KCl + 2MnCl2 + 5Cl2 + 8H2O
2a mol -> 16a mol -> 5a mol
MnO2 + 4HCl -> MnCl2 + Cl2 + 2H2O
b mol -> 4b mol -> b mol
Ta co : bao toan khoi luong hon hop truoc phan ung :
158*2a + 87b =35,88 (1)
bao toan so mol Cl2 sinh ra :
5a + b = 0,515 (2)
giai he pt (1) va (2) ,ta duoc : a = 0,075 mol , b =0,14 mol
-> mKMnO4 pu = 0,075*2*158 = 23,7 gam
-> mMnO2 pu = 35,88 - 23,7 = 12,18 gam
b, tong so mol HCl tham gia phan ung la :
16a + 4b = 16*0,075 + 4*0,14 = 1,76 mol
-> mct HCl = 1,76 *36,5 = 64,24 gam
ta co : C% = mct/mdd*100%
-> mdd HCl= mct*100%/C% = 64,24*100%/37% = 173,6216 gam