a,ta co pthh
Zn + 2HCl \(\rightarrow\) ZnCl2 + H2
b,Theo de bai ta co
nZn=\(\dfrac{29,25}{65}=0,45mol\)
b,Theo pthh
nH2=nZn=0,45 mol
\(\Rightarrow VH2_{\left(dktc\right)}=0,45.22,4=10,08l\)
c,Theo pthh
nZnCl2=nZn=0,45 mol
\(\Rightarrow\) mZnCl2=0,45.136=61,2g
d,Theo pthh
nHCl=2nZn=2.0,45=0,9mol
\(\Rightarrow VddHCl=\dfrac{nct}{CM}=\dfrac{0,9}{2,5}\)=0,36l=360 ml