Em tham khảo nha:

\(\begin{array}{l}
V{D_1}:\\
{n_{{C_2}{H_2}}} = \dfrac{N}{{6 \times {{10}^{23}}}} = \dfrac{{7,5 \times {{10}^{30}}}}{{6 \times {{10}^{23}}}} = 12500000\,mol\\
{m_{{C_2}{H_2}}} = 12500000 \times 26 = 325000000g = 325\text{ tấn}\\
{V_{{C_2}{H_2}}} = 12500000 \times 22,4 = 280000000l = 280000{m^3}\\
V{D_2}:\\
{n_{{C_2}{H_4}}} = \dfrac{V}{{22,4}} = \dfrac{{21280}}{{22,4}} = 950\,mol\\
{m_{{C_2}{H_4}}} = n \times M = 950 \times 28 = 26600g\\
{N_{{C_2}{H_4}}} = 950 \times 6 \times {10^{23}} = 5,7 \times {10^{26}}\text{ phân tử}\\
V{D_3}:\\
{n_{C{H_4}}} = \dfrac{m}{M} = \dfrac{{1280}}{{16}} = 80\,mol\\
{V_{C{H_4}}} = 80 \times 22,4 = 1792l\\
{N_{C{H_4}}} = 80 \times 6 \times {10^{23}} = 4,8 \times {10^{25}}\text{ phân tử}
\end{array}\)

Đáp án:

Giải thích các bước giải: