\(n_A=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(V\%_{C2H4}=\frac{0,84}{3,36}.100\%=25\%\)
\(\Rightarrow V\%_{C3H4}=100\%-25\%=75\%\)
\(n\%_{C3H4}=V\%_{C3H4}=75\%\)
\(\Rightarrow n\%_{C3H4}=0,015.75\%=0,1125\)
\(n_{C3H4}=n_{C3H4Ag}=0,1125\left(mol\right)\)
\(\Rightarrow m_{kt}=0,1125.127=16,53\left(g\right)\)