Câu $1$$HCHO+4AgNO_3+6NH_3+2H_2O\rightarrow (NH_4)_2CO_3+4Ag\downarrow+4NH_4NO_3$
$\Rightarrow n_{Ag}=4n_{HCHO}=\frac{4.6}{30}=0,8 mol\Rightarrow m=86,4 gam$
Câu $2$
$2H_2+CH_2=CH-CHO\rightarrow CH_3CH_2CH_2OH$
$\Rightarrow n_{H_2}=\frac{2.11,2}{56}=0,4 mol\Rightarrow V_{dkc}=8,96 lit\Rightarrow V=\frac{8,96}{2}=4,48 lit$