Question

Answer 1

5)

nCO2=2,24/22,4=0,1(mol)

PTHH: Ca(OH)2 + CO2 -> CaCO3 + H2O

0,1________0,1_______0,1(mol)

CMddCa(OH)2= 0,1/0,2=0,5(M)

mCaCO3=0,1.100=10(g)

Answer 2

6)

PTHH: 2 NaOH + CuCl2 -> Cu(OH)2 + 2 NaCl

Ta có: 0,5/2 > 0,2/1

-> CuCl2 hết, NaOH dư -> tính theo nCuCl2

PTHH: Cu(OH)2 -to-> CuO + H2O

nCuO= nCu(OH)2= nCuCl2= 0,2(mol)

nNaCl=nNaOH(p.ứ)=2.0,2=0,4(mol)

-> M(rắn) =  mCuO =  0,2.80= 16(g)

B) mddsau= 200+319,6 - 0,2.98= 500(g)

mNaOH(dư)= (0,5-0,2.2).40=4(g)

mNaCl=0,4.58,5=23,4(g)

C%ddNaOH(dư)= (4/500).100= 0,8%

C%ddNaCl= (23,4/500).100=4,68%