Ta co pthh
Zn + 2HCl \(\rightarrow\)ZnCl2 + H2
Theo de bai ta co
nZn=\(\dfrac{1,95}{65}=0,03mol\)
Theo pthh
nH2 = nZn= 0,03 mol
\(\Rightarrow\)VH2 = 0,03 .22,4 = 0,672 l
theo pthh
nZnCl2 = nZn =0,03 mol
\(\Rightarrow\)mZnCl2 = 0,03 .136=4,08 g
mH2 = 0,03 .2 = 0,06 g
\(\Rightarrow\) Khoi luong cua dd sau phan ung la
mdd = 1,95 + 120 -0,06 = 121,89 g
\(\Rightarrow C\%=\dfrac{4,08}{121,89}.100\%\approx3,35\%\)