\(d_{A/CH4}=\frac{17}{32}\rightarrow\overline{M_A}=\frac{17.16}{32}=8,5\left(\frac{g}{mol}\right)\)
\(\rightarrow\overline{M_A}=\frac{2n_{H2}+28n_{CO}}{n_{H2}+n_{CO}}=8,5\left(\frac{g}{mol}\right)\)
\(\rightarrow2n_{H2}+28n_{CO}=8,5n_{H2}+8,5n_{CO}\)
\(\rightarrow19,5n_{CO}=6,5n_{H2}\rightarrow3n_{CO}=n_{H2}\)
Gọi số mol của H2 và CO lần lượt là a ; b ( a ; b> 0 )
\(\rightarrow3a=a\rightarrow a-3b=0\left(1\right)\)
PTHH:
\(2H_2+O_2\rightarrow2H_2O\)
a____0,5a__________
\(2CO+O_2\rightarrow2CO_2\)
b______0,5b______
\(\rightarrow0,5a+0,5b=\frac{4,48}{22,4}=0,2\left(mol\right)\left(2\right)\)
Từ (1) và (2)
\(\rightarrow\left\{{}\begin{matrix}a-3b=0\\0,5a+0,5b=0,2\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}a=0,3\\b=0,1\end{matrix}\right.\)
\(\rightarrow\%V_{H2}=\frac{0,3}{0,3+0,1}.100\%=75\%\)
\(\rightarrow\%V_{CO}=100\%-75\%=25\%\)
