$NaOH + HCl \to NaCl + H_2O$
Theo PTHH :
$n_{HCl} = n_{NaOH} = 0,2.1 = 0,2(mol)$
$m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)$
`n_{NaOH} = 0,2.1 = 0,2 (mol)`
PTHH: `NaOH + HCl -> NaCl + H_2O`
Theo PT: `n_{HCl} = n_{NaOH} = 0,2 (mol)`
`=> m = (0,2.36,5)/(7,3\%) = 100 (g)`