Đáp án:

`75\%` 

Giải thích các bước giải:

`n_{Ag}={16,2}/{108}=0,15(mol)`

`n_X={2,479}/{24,79}=0,1(mol)`

`C_2H_4+1/2O_2` $\xrightarrow{t^o,xt}$ `CH_3CHO`

`CH_3CHO+2[Ag(NH_3)_2]OH->CH_3COONH_4+2Ag+3NH_3+H_2O`

Theo PT: `n_{CH_3CHO}=1/2n_{Ag}=0,075(mol)`

`->n_{C_2H_4\ du}=0,1-0,075=0,025(mol)`

Theo PT: `n_{C_2H_4\ pu}=n_{CH_3CHO}=0,075(mol)`

`->H={0,075}/{0,075+0,025}.100\%=75\%`