a) \(n_A=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)
\(\overline{M}_A=15.2=30\left(g/mol\right)\)
Áp dụng sơ đồ đường chéo, ta có:
\(\dfrac{n_{CO}}{n_{O_2}}=\dfrac{32-30}{30-28}=\dfrac{1}{1}\)
=> \(n_{CO}=n_{O_2}=\dfrac{0,4}{2}=0,2\left(mol\right)\)
=> \(A\left\{{}\begin{matrix}\%V_{CO}=\dfrac{0,2}{0,4}.100\%=50\%\\\%V_{O_2}=100\%-50\%=50\%\end{matrix}\right.\)
PTHH: \(2CO+O_2\xrightarrow[]{t^o}2CO_2\)
ban đầu 0,2 0,2
phản ứng 0,2--->0,1
sau phản ứng 0 0,1 0,2
=> \(B\left\{{}\begin{matrix}\%V_{O_2}=\dfrac{0,1}{0,1+0,2}.100\%=33,33\%\\\%V_{CO_2}=100\%-33,33\%=66,67\%\end{matrix}\right.\)
b) Theo ĐLBTKL: \(m_A=m_B\)
c) \(\overline{M}_B=\dfrac{0,1.32+0,2.44}{0,1+0,2}=40\left(g/mol\right)\)
=> \(d_{B/H_2}=\dfrac{40}{2}=20\)
