Question

Answer 1

nHCl=0,2.0,2=0,04(mol)

a) PTHH: HCl + NaOH -> NaCl + H2O

0,04________0,04____0,04(mol)

=>VddNaOH= 0,04/0,1=0,4(l)=400(ml)

Vddsau=VddNaOH + VddHCl=400+200=600(ml)=0,6(l)

=>CMddNaOH= 0,04/0,6= 1/15(M)

b) Ca(OH)2 + 2 HCl -> CaCl2 + H2O

0,02_________0,04___0,02(mol)

=>mCa(OH)2=0,02.74=1,48(g)

=>mddCa(OH)2= (1,48.100)/5= 29,6(g)

mddsau= 200.1+ 29,6= 209,6(g)

mCaCl2= 0,02.111=2,22(g)

=>C%ddCaCl2=(2,22/229,6).100=0,967%