Em tham khảo nha:

\(\begin{array}{l}
7)\\
a)\\
{m_{{H_2}S{O_4}}} = n \times M = 2 \times 98 = 198g\\
b)\\
{n_{S{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
{m_{S{O_2}}} = n \times M = 0,3 \times 64 = 19,2g\\
c)\\
{n_{Zn}} = \dfrac{N}{{6 \times {{10}^{23}}}} = \dfrac{{3 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,5\,mol\\
{m_{Zn}} = n \times M = 0,5 \times 65 = 32,5g\\
d)\\
{m_{hh}} = {m_{{O_2}}} + {m_{{N_2}}} = 0,2 \times 32 + 0,8 \times 28 = 28,8g\\
e)\\
{n_{{O_2}}} = \dfrac{N}{N} = 1\,mol\\
{n_{C{H_4}}} = \dfrac{{2N}}{N} = 2\,mol\\
m = 1 \times 32 + 2 \times 16 = 64g\\
8)\\
a)\\
{n_{Na}} = \dfrac{m}{M} = \dfrac{{4,6}}{{23}} = 0,2\,mol\\
b)\\
{n_{C{O_2}}} = \dfrac{{13,44}}{{22,4}} = 0,6mol\\
c)\\
{n_{NaOH}} = \dfrac{{12 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 2\,mol\\
9)\\
a)\\
{N_{S{O_2}}} = 0,25 \times 6 \times {10^{23}} = 1,5 \times {10^{23}}\\
b)\\
{n_{N{H_3}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{N_{N{H_3}}} = 0,2 \times 6 \times {10^{23}} = 1,2 \times {10^{23}}\\
c)\\
{n_{{O_2}}} = \dfrac{{6,4}}{{32}} = 0,2\,mol\\
{N_{{O_2}}} = 0,2 \times 6 \times {10^{23}} = 1,2 \times {10^{23}}
\end{array}\)

Đáp án:

Giải thích các bước giải: