a, Ta có: \(n_{Ag_2C_2}=\dfrac{43,2}{240}=0,18\left(mol\right)\)
BTNT C, có: \(n_{C_2H_2}=n_{Ag_2C_2}=0,18\left(mol\right)\)
Mà: \(n_{hhk}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow n_{C_3H_8}=0,07\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,18}{0,25}.100\%=72\%\\\%V_{C_3H_8}=28\%\end{matrix}\right.\)
và \(\left\{{}\begin{matrix}\%m_{C_2H_2}=\dfrac{0,18.26}{0,18.26+0,07.44}.100\%\approx60,3\%\\\%m_{C_3H_8}=39,7\%\end{matrix}\right.\)
b, BTNT C, có: \(n_{CO_2}=3n_{C_3H_8}=0,21\left(mol\right)\)
\(\Rightarrow m_{CO_2}=0,21.44=9,24\left(g\right)\)
Bạn tham khảo nhé!
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