A)ZnO + 2HCl -> ZnCl_2 + H_2O

n_{ZnO} = {24,3}/{81} = 0,3 mol

B)n_{HCl} = 2n_{ZnO} = 2 . 0,3 = 0,6 mol

=> C%_{dd HCl} = {0,6 . 36,5}/{500} . 100 = 4,38%

c)m_{dd sau} = 24,3+ 500 = 524,3g

n_{ZnCl_2} = n_{ZnO} = 0,3 mol

=> C%_{dd ZnCl_2} = {0,3 . 136}/{524,3} . 100% = 7,782%