`nZnO=(4,05)/(81)=0,05 (mol)`
`mHCl = 100.3,65%=3,65(g)`
` => nHCl = (3,65)/(36,5)=0,1 (mol)`
a,`ZnO + 2HCl -> ZnCl2 + H2O`
b,
`=> 2.nZnO=nHCl (0,05.2=0,1)`
`=>` pứ vừa đủ ko dư
c,
`nZnCl2= nZnO=0,05 (mol)`
` => mZnCl2=0,05.136=6,8 (g)`
`m_(dd)= mZnO + m_(dd_(HCl))`
`= 4,05 + 100=104,5 (g)`
`=> %ZnCl2= (6,8)/(104,5).100=6,5%`