Question

Answer 1

8)

nCaCl2=0,03.0,5=0,015(mol); nAgNO3=0,07.1=0,07(mol)

PTHH: CaCl2 + 2 AgNO3 -> Ca(NO3)2 + 2 AgCl

Ta có: 0,07/2> 0,015/1

-> AgNO3 dư, CaCl2 hết => Tính theo  nCaCl2

=> nAgCl= 2.nCaCl2=0,015.2=0,03(mol)

=>m(kết tủa)= mAgCl= 0,03.143,5=4,305(g)

nAgNO3(dư)=0,07- 0,015.2= 0,04(mol)

nCa(NO3)2= nCaCl2= 0,015(mol)

Vddsau= 30+70=100(ml)=0,1(l)

=>CMddAgNO3(DƯ)= 0,04/0,1=0,4(M)

CMddCa(NO3)2= 0,015/0,1=0,15(M)

Answer 2

Câu 9:

mNa2CO3=265.10%=26,5(g) => nNa2CO3= 26,5/106=0,25(mol)

mCaCl2=6,66%. 500=33,3(g) => nCaCl2= 33,3/111=0,3(mol)

PTHH: Na2CO3 + CaCl2 -> CaCO3 (KT) + 2 NaCl

Ta có: 0,3/1 > 0,25/1

-> Na2CO3 hết, CaCl2 dư => Tính theo nNa2CO3.

=> nCaCl2(P.ứ)= nNa2CO3=0,25(mol)

=> nCaCl2(dư)=0,3-0,25=0,05(mol) => mCaCl2(dư)= 0,05.111= 5,55(g)

nNaCl=0,25.2=0,5(mol) => mNaCl=0,5.58,5=29,25(g)

mddsau= 265+500 - 0,25.100= 740(g)

=> C%ddCaCl2(dư)= (5,55/740).100=0,75%

C%ddNaCl= (29,25/740).100=3,953%