Em tham khảo nha:
\(\begin{array}{l}
1)\\
{n_{C{O_2}}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{n_C} = {n_{C{O_2}}} = 0,3\,mol \Rightarrow {m_C} = 0,3 \times 12 = 3,6g\\
{n_O} = 2{n_{C{O_2}}} = 0,6\,mol \Rightarrow {m_O} = 0,6 \times 16 = 9,6g\\
2)\\
{n_{F{e_2}{{(S{O_4})}_3}}} = \dfrac{{100}}{{400}} = 0,25\,mol\\
{n_{Fe}} = 2{n_{F{e_2}{{(S{O_4})}_3}}} = 0,5\,mol \Rightarrow {m_{Fe}} = 0,5 \times 56 = 28g\\
{n_S} = 3{n_{F{e_2}{{(S{O_4})}_3}}} = 0,75\,mol \Rightarrow {m_S} = 0,75 \times 32 = 24g\\
{m_O} = 100 - 28 - 24 = 48g
\end{array}\)
\(\begin{array}{l}
A:{(N{H_2})_2}CO\\
{(N{H_2})_2}CO + 2{H_2}O \to {(N{H_4})_2}C{O_3}\\
B:{(N{H_4})_2}C{O_3}\\
{(N{H_4})_2}C{O_3} + 2NaOH \to N{a_2}C{O_3} + 2N{H_3} + 2{H_2}O\\
D:N{a_2}C{O_3}\\
E:N{H_3}\\
{(N{H_4})_2}C{O_3} + 2HCl \to 2N{H_4}Cl + C{O_2} + {H_2}O\\
G:N{H_4}Cl\\
H:C{O_2}\\
{(N{H_4})_2}C{O_3} + C{O_2} + {H_2}O \to 2N{H_4}HC{O_3}\\
L:N{H_4}HC{O_3}\\
N{H_4}HC{O_3} \xrightarrow{t^0} N{H_3} + C{O_2} + {H_2}O
\end{array}\)
