Đáp án:

 a) 2,4 mol

b) 0,4167mol

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
{M_A} = 19,2 \times 2 = 38,4g/mol\\
{m_A} = 38,4 \times 1 = 38,4g\\
hh:{O_2}(a\,mol),{O_3}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 1\\
32a + 48b = 38,4
\end{array} \right.\\
 \Rightarrow a = 0,6;b = 0,4\\
2CO + {O_2} \xrightarrow{t^0} 2C{O_2}\\
3CO + {O_3} \xrightarrow{t^0}3C{O_2}\\
{n_{CO}} = 0,6 \times 2 + 0,4 \times 3 = 2,4\,mol\\
b)\\
{M_B} = 3,6 \times 2 = 7,2\,g/mol\\
{m_B} = 7,2 \times 1 = 7,2g\\
hh\,B:CO(a\,mol),{H_2}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 1\\
28a + 2b = 7,2
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,8\\
2{H_2} + {O_2} \xrightarrow{t^0} 2{H_2}O\\
2CO + {O_2} \xrightarrow{t^0}2C{O_2}\\
3{H_2} + {O_3} \xrightarrow{t^0} 3{H_2}O\\
3CO + {O_3} \xrightarrow{t^0} 3C{O_2}\\
{n_{{H_2}O}} = {n_{{H_2}}} = 0,8\,mol\\
{n_{C{O_2}}} = {n_{CO}} = 0,2\,mol\\
BTKL:\\
{m_A} + {m_B} = {m_{C{O_2}}} + {m_{{H_2}O}}\\
 \Rightarrow {m_A} + 7,2 = 0,8 \times 18 + 0,2 \times 44\\
 \Rightarrow {m_A} = 16g \Rightarrow {n_A} = \dfrac{{16}}{{38,4}} = \dfrac{5}{{12}} \approx 0,4167\,mol
\end{array}\)