nZn=0.2 mol
Pthh
Zn +2HCl -----> ZnCl2 +H2
0.2------------>0.2 mol
=>nZnCl2=nZn= 0.2 mol
=>mZnCl2= 0.2*136 =27.2g
Zn + 2HCl----->ZnCl2 +H2
Ta có
n\(_{Zn}=\frac{13}{65}=0,2\left(mol\right)\)
Theo pthh
n\(_{ZnCl}=n_{Zn}=0,2\left(mol\right)\)
mdd =200+13 -0,4=212,6(g)
C%(ZnCl2)=\(\frac{0,2.136}{212,6}.100\%=12,79\%\)
Chúc bạn học tốt