\(n_{CaCO_3^{}}^{} = \frac{{0,25}}{{100}} = 0,0025\;{\text{mol}}\)

\(Ca(OH)_2^{} + CO_2^{} \to CaCO_3^{} + H_2^{}O\)

\({\text{                0}}{\text{,0025         0}}{\text{,0025}}\)

\(CH_4^{} + 2O_2^{}\xrightarrow{{{t^o}}}CO_2^{} + 2H_2^{}O\)

\(\frac{{0,0025}}{1} < \frac{{0,01}}{1} \to {\text{O}}_2^{}\;{\text{du}}{\text{, tinh theo CH}}_4^{}\)

\(0,0025{\text{  0}}{\text{,005       0}}{\text{,0025}}\)\

\(a = 0,0025.16 = 0,04g\)

\(N_{O_{2_{du}^{}}^{}}^{} = 6,022.10_{}^{23}.(0,01 - 0,005) = {3,011.10^{21}}(NguyenTu)\)

\(\begin{array}{l}
a)\\
n_{O_2}=\frac{0,32}{32}=0,01(mol)\\
n_{CaCO_3}=\frac{0,25}{100}=0,0025(mol)\\
CH_4+2O_2\xrightarrow{t^o}CO_2+2H_2O\\
CO_2+Ca(OH)_2\to CaCO_3+H_2O\\
Theo\,PT:\,n_{CH_4}=n_{CO_2}=n_{CaCO_3}=0,0025(mol)\\
\to a=0,0025.6.10^{23}=1,5.10^{21}(\rm phan\,tu)\\
b)\\
Theo\,PT:\,n_{O_2(pu)}=2n_{CH_4}=0,005(mol)\\
\to n_{O_2(du)}=0,01-0,005=0,005(mol)\\
\to A_{O_2(du)}=0,005.6.10^{23}=3.10^{21}(\rm phan\,tu)
\end{array}\)