Question

Mình cảm ơn❤️

Answer 1

$n_{HCl} = \dfrac{350.14,6\%}{36,5} = 1,4(mol)$

$KOH + HCl \to KCl + H_2O$

$n_{HCl\ dư} = n_{KOH} = 0,14.2 = 0,28(mol)$
$\Rightarrow n_{HCl\ pư} = 1,4 - 0,28 = 1,12(mol)$

$O_{oxit} + 2H_{axit} \to H_2O$

Ta có : $n_H = n_{HCl} = 1,12(mol) \Rightarrow n_{O\ trong\ oxit} = \dfrac{1}{2}n_H = 0,56(mol)$
$CO + O_{oxit} \to CO_2$
$n_{CO} = n_O = 0,56(mol)$
$\Rightarrow V = 0,56.22,4 = 12,544(lít)$