$HCHO+4AgNO_3+2H_2O+6NH_3\xrightarrow{t^o}(NH_4)_2CO_3+4Ag+4NH_4NO_3$
x->4x (mol)
$C_2H_5CHO+2AgNO_3+H_2O+3NH_3\xrightarrow{}C_2H_5COONH_4+2Ag+2NH_4NO_3$
y -> 2y (mol)
$n_{Ag}=\dfrac{34,56}{108}=0,32(mol)=4x+2y(1)$
$m_{hh}=30x+58y=3,8(g)(2)$
$(1),(2)=>x=\dfrac{137}{2150}(mol);y=\dfrac{7}{215}(mol)$
=> %$m_{HCHO}=\dfrac{\dfrac{137}{2150}.30}{3,8}.100≈50,31$%
=> ko có đáp án thỏa đề bài !
