\(a,n_{kết.tủa}=n_{Ag_2C_2}=\dfrac{24}{240}=0,1\left(mol\right)\\ V_{hh}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH: \(C_2H_2+AgNO_3+NH_3\rightarrow Ag_2C_2\downarrow+2NH_4NO_3\)
0,1<----------------------------0,1
\(m_{bình.Br_2.tăng}=m_{C_2H_2}\\ \rightarrow n_{C_2H_2}=\dfrac{5,6}{28}=0,2\left(mol\right)\\ \rightarrow n_{C_2H_6}=0,5-0,1-0,2=0,2\left(mol\right)\\ \rightarrow\left\{{}\begin{matrix}\%V_{C_2H_2}=\dfrac{0,1}{0,5}.100\%=20\%\\\%V_{C_2H_4}=\dfrac{0,2}{0,5}.100\%=40\%\\\%V_{C_2H_6}=100\%-20\%-40\%=40\%\end{matrix}\right.\)
\(b,m_{hh}=0,1.26+0,2.28+0,2.30=14,2\left(g\right)\\ \rightarrow M_{hh}=\dfrac{14,2}{0,5}=28,4\left(g\text{/}mol\right)\\ \rightarrow d_{hh\text{/}kk}=\dfrac{28,4}{29}=0,98\)
