Em tham khảo nha:

\(\begin{array}{l}
a)\\
{n_{Mg{{(OH)}_2}}} = \dfrac{N}{{6 \times {{10}^{23}}}} = \dfrac{{3 \times {{10}^{23}}}}{{6 \times {{10}^{23}}}} = 0,5\,mol\\
{m_{Mg{{(OH)}_2}}} = n \times M = 0,5 \times 58 = 29g\\
{n_{{H_2}}} = \dfrac{V}{{22,4}} = \dfrac{{11,2}}{{22,4}} = 0,5\,mol\\
{m_{{H_2}}} = n \times M = 0,5 \times 2 = 1g\\
{m_{N{a_3}P{O_4}}} = n \times M = 0,15 \times 164 = 24,6g\\
b)\\
{n_{S{O_3}}} = \dfrac{m}{M} = \dfrac{{49}}{{80}} = 0,6125\,mol\\
{V_{S{O_3}}} = n \times 22,4 = 0,6125 \times 22,4 = 13,72l\\
{n_{{O_2}}} = \dfrac{m}{M} = \dfrac{{0,8}}{{32}} = 0,025\,mol\\
{V_{{O_2}}} = n \times 22,4 = 0,025 \times 22,4 = 0,56l\\
{n_{C{H_4}}} = \dfrac{{1,2N}}{N} = 1,2\,mol\\
{V_{C{H_4}}} = n \times 22,4 = 1,2 \times 22,4 = 26,88l\\
c)\\
{N_{HCl}} = n \times 6 \times {10^{23}} = 0,6 \times 6 \times {10^{23}} = 3,6 \times {10^{23}}
\end{array}\)

Đáp án:

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