\(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
a,\(V_{C2H6}=2,24\left(l\right)\)
\(V_{C2H4}=6,72-22,4=4,48\left(l\right)\)
\(V\%_{C2H4}=\frac{2,24}{6,72}.100\%=33,33\%\)
\(V\%_{C2H4}=100\%-33,33\%=66,67\%\)
b,\(n_{C2H4}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
\(\Rightarrow m=m_{C2H4}=0,2.28=5,6\left(g\right)\)
\(n_{NaOH}=0,1.1,25=0,125\left(mol\right)\)
\(n_{H2S}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(2NaOH+H_2S\rightarrow Na_2S+2H_2O\left(1\right)\)
\(Na_2S+H_2S\rightarrow2NaHS\left(2\right)\)
\(n_{H2S\left(1\right)}=n_{Na2S\left(1\right)}=0,0625\left(mol\right)\)
\(n_{H2S\left(2\right)}=n_{Na2S\left(2\right)}=0,1-0,0625=0,0375\left(mol\right)\)
\(n_{NaHS}=0,0375.2=0,075\left(mol\right)\)
\(n_{Na2S}=0,0625-0,0375=0,025\left(mol\right)\)
\(m_{Na2S}=0,025.78=1,95\left(g\right)\)
\(\Rightarrow m_{NaHS}=0,075.56=4,2\left(g\right)\)