Em tham khảo nha :

\(\begin{array}{l}
5)\\
3CO + F{e_2}{O_3} \to 2Fe + 3C{O_2}\\
a)\\
{n_{CO}} = \dfrac{{22,4}}{{22,4}} = 1mol\\
{n_{F{e_2}{O_3}}} = \dfrac{1}{3}{n_{CO}} = \dfrac{1}{3}mol\\
{m_{F{e_2}{O_3}}} = n \times M = \dfrac{1}{3} \times 160 = 53,3g\\
b)\\
{n_{Fe}} = \dfrac{2}{3}{n_{CO}} = \dfrac{2}{3}mol\\
{m_{Fe}} = n \times M = \dfrac{2}{3} \times 56 = 37,3g\\
c)\\
{n_{C{O_2}}} = {n_{CO}} = 1mol\\
{V_{C{O_2}}} = n \times 22,4 = 1 \times 22,4 = 22,4l\\
6)\\
a)\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
b)\\
{n_{Na}} = \dfrac{m}{M} = \dfrac{{11,5}}{{23}} = 0,5mol\\
{n_{{H_2}}} = \dfrac{{{n_{Na}}}}{2} = 0,25mol\\
{V_{{H_2}}} = n \times 22,4 = 0,25 \times 22,4 = 5,6l\\
c)\\
{n_{NaOH}} = {n_{Na}} = 0,5mol\\
{m_{NaOH}} = n \times M = 0,5 \times 40 = 20g\\
7)\\
a)\\
Fe + 3AgN{O_3} \to 3Ag + Fe{(N{O_3})_3}\\
b)\\
{n_{Ag}} = \dfrac{m}{M} = \dfrac{{10,8}}{{108}} = 0,1mol\\
{n_{Fe}} = \dfrac{{{n_{Ag}}}}{3} = \frac{1}{{30}}mol\\
{m_{Fe}} = n \times M = \dfrac{1}{{30}} \times 56 = 1,867g\\
{n_{AgN{O_3}}} = {n_{Ag}} = 0,1mol\\
{m_{AgN{O_3}}} = n \times M = 0,1 \times 170 = 17g\\
c)\\
{n_{Fe{{(N{O_3})}_3}}} = {n_{Fe}} = \dfrac{1}{{30}}mol\\
{m_{Fe{{(N{O_3})}_3}}} = n \times M = \dfrac{1}{{30}} \times 242 = 8,067g
\end{array}\)

Đáp án:

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