Xét 4 gam \(NaOH\).
\({n_{NaOH}} = \frac{4}{{23 + 16 + 1}} = 0,1{\text{ mol = }}{{\text{n}}_{Na}} = {n_O} = {n_H}\)
\( \to {m_{Na}} = 0,1.23 = 2,3{\text{ gam}}\)
\({m_O} = 0,1.16 = 1,6{\text{ gam}}\)
\({m_H} = 0,1.1 = 0,1{\text{ gam}}\)
Xét 32 gam \(Fe_2(SO_4)_3\).
Ta có:
\({n_{F{e_2}{{(S{O_4})}_3}}} = \frac{{32}}{{56.2 + (32 + 16.4).3}} = 0,08{\text{ mol}}\)
\( \to {n_{Fe}} = 2{n_{F{e_2}{{(S{O_4})}_3}}} = 0,08.2 = 0,16{\text{ mol}}\)
\({n_S} = 3{n_{F{e_2}{{(S{O_4})}_3}}} = 0,24{\text{ mol}}\)
\({n_O} = 12{n_{F{e_2}{{(S{O_4})}_3}}} = 0,96{\text{ mol}}\)
\( \to {m_{Fe}} = 0,16.56 = 8,96{\text{ gam}}\)
\( \to {m_S} = 0,24.32 = 7,68{\text{ gam}}\)
\( \to {m_O} = 0,96.16 = 15,36{\text{ gam}}\)
