a/
\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
\(C_2H_2+2Br_2\rightarrow C_2H_2Br_4\)
\(n_{Br_2}=\frac{48}{160}=0,3\left(mol\right)=n_{C_2H_4}+2n_{C_2H_2}\left(1\right)\)
\(C_2H_2+2AgNO_3+2NH_3\rightarrow Ag_2C_2\downarrow+2NH_4NO_3\)
b/
\(n_{C_2H_2}=n_{Ag_2C_2}=\frac{24}{240}=0,1\left(mol\right)\)
thay n\(_{C_2H_2}\) vào (1) ta có:
\(n_{C_2H_4}+2.0,1=0,3\)
\(\Leftrightarrow n_{C_2H_4}=0,3-0,2=0,1\left(mol\right)\)
\(m_{C_2H_2}=0,1.26=2,6\left(g\right)\)
\(m_{C_2H_4}=0,1.28=2,8\left(g\right)\)
\(n_{CH_4}=7-2,6-2,8=1,6\left(g\right)\)
Đúng 0
Bình luận (0)
