Em tham khảo nha :

\(\begin{array}{l}
10)\\
a)\\
{n_{{O_2}}} = \dfrac{{{N_O}}}{{{N_A}}} = \frac{N}{N} = 1mol\\
{m_{{O_2}}} = n \times M = 1 \times 32 = 32g\\
{n_{{N_2}}} = \dfrac{{{N_{{N_2}}}}}{{{N_A}}} = \frac{{2N}}{N} = 2mol\\
{m_{{N_2}}} = n \times M = 2 \times 28 = 56g\\
{n_{C{O_2}}} = \dfrac{{{N_{C{O_2}}}}}{{{N_A}}} = \frac{{1,5N}}{N} = 1,5mol\\
{m_{C{O_2}}} = n \times M = 1,5 \times 44 = 66g\\
m = {m_{{O_2}}} + {m_{{N_2}}} + {m_{C{O_2}}} = 32 + 56 + 66 = 154g\\
b)\\
{m_{Fe}} = n \times M = 0,1 \times 56 = 5,6g\\
{m_{Cu}} = n \times M = 0,2 \times 64 = 12,8g\\
{m_{Zn}} = n \times M = 0,3 \times 65 = 19,5g\\
{m_{Al}} = n \times M = 0,25 \times 27 = 6,75g\\
m = {m_{Fe}} + {m_{Cu}} + {m_{Zn}} + {m_{Al}} = 5,6 + 12,8 + 19,5 + 6,75 = 44,65g\\
c)\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{m_{{O_2}}} = n \times M = 0,1 \times 32 = 3,2g\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
{m_{{H_2}}} = n \times M = 0,05 \times 2 = 0,1g\\
{n_{HCl}} = \dfrac{V}{{22,4}} = \dfrac{{6,72}}{{22,4}} = 0,3mol\\
{m_{HCl}} = n \times M = 0,3 \times 36,5 = 10,95g\\
{n_{C{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{0,56}}{{22,4}} = 0,025mol\\
{m_{C{O_2}}} = n \times M = 0,025 \times 44 = 1,1g\\
m = {m_{{O_2}}} + {m_{{H_2}}} + {m_{HCl}} + {m_{C{O_2}}} = 3,2 + 0,1 + 10,95 + 1,1 = 15,35g
\end{array}\)