Câu 27:
\(C_3H_4+AgNO_3+NH_3\rightarrow C_3H_3Ag+NH_4NO_3\\ C_2H_2+2AgNO_3+2NH_3\rightarrow Ag_2C_2+2NH_4NO_3\\ n_{C_3H_4}=\dfrac{1,8}{40}=0,045\left(mol\right);n_{C_2H_2}=\dfrac{0,52}{26}=0,02\left(mol\right)\\ n_{C_3H_3Ag}=n_{C_3H_4}=0,045\left(mol\right)\\ n_{Ag_2C_2}=n_{C_2H_2}=0,02\left(mol\right)\\ m_{kt}=m_{C_3H_3Ag}+m_{Ag_2C_2}=0,045.147+240.0,02=11,415\left(g\right)\\ Chọn.B\)
\(n_{Br_2}=\dfrac{32}{160}=0,2\left(mol\right)\\ Đặt:n_{C_3H_6}=a\left(mol\right);n_{C_3H_4}=b\left(mol\right)\\ PTHH:C_3H_6+Br_2\rightarrow C_3H_6Br_2\\ C_3H_4+2Br_2\rightarrow C_3H_4Br_4\\ \Rightarrow\left\{{}\begin{matrix}42a+40b=6,2\\a+2b=0,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \Rightarrow\%m_{C_3H_6}=\dfrac{0,1.42}{6,2}.100\approx67,74\%\\ Chọn.A\)
\(n_{Br_2}=\dfrac{0,8}{160}=0,005\left(mol\right)\\ C_2H_4\left(dư\right)+Br_2\rightarrow C_2H_4Br_2\\ n_{C_2H_4\left(dư\right)}=n_{Br_2}=0,005\left(mol\right)\\ n_{C_2H_4\left(đã.p.ứ.với.H_2\right)}=\dfrac{2,24}{22,4}-0,005=0,095\left(mol\right)\\ C_2H_4+H_2\rightarrow\left(Ni,t^o\right)C_2H_6\\ Vì:\dfrac{0,095}{1}< \dfrac{3,36:22,4}{1}\Rightarrow C_2H_4hết,H_2dư\\ H=\dfrac{0,095}{0,1}.100\%=95\%\\ Chọn.C\)
