`nZn =0,1 (mol)`
` Zn + 2HCl -> ZnCl_2 + H_2`
`a, nH_2 = nZnCl_2=nZn = 0,1 (mol)`
`=> V_(H_2) = 2,24 (l)`
` => mZnCl_2 =13,6 (g`)
b,
`mdd` mới `= mZn + mdd_(HCl) - mH_2`
`= 6,5 + 500-0,1.2`
`=506,3 (g)`
` => % ZnCl_2 =(13,6)/(506,3).100=2,68%`
c,
`ZnCl_2 + 2KOH -> 2KCl + Zn(OH)_2`
` nZn(OH)2 = nZnCl2 =0,1 (mol)`
`=> mZn(OH)2=9,9(g)`
` nKOH = 2.nZnCl2 = 0,2 (mol)`
`=> V_(KOH) = 0,2:2=0,1 (l)`