Đáp án: `151,2g`

Lời giải:

Glucozo: `C_6H_12O_6:3xmol`

Saccarozo: `C_12H_22O_11:2xmol`

`=>180.3x+342.2x=122,4`

`<=>x=0,1`

`=>n_{C_6H_12O_6}=0,3mol`

`n_{C_12H_22O_11}=0,2mol`

$C_{12}H_{22}O_{11}+H_2O\xrightarrow[t^o]{H_2SO_4}2C_6H_{12}O_6$

`=>n_{C_6H_12O_6(Y)}=0,2.2+0,3=0,7(mol)`

Vì trong `C_6H_12O_6` có `1` nhóm `CHO`

`1CHO->2Ag`

`=>n_{Ag}=0,7.2=1,4(mol)`

`=>m_{Ag}=1,4.108=151,2(g)`

`n_{C_6H_{12}O_6}=3x(mol);n_{C_{12}H_{22}O_{11}}=2x(mol)`

`->180.3x+342.2x=122,4`

`->x=0,1(mol)`

`->n_{C_6H_{12}O_6}=0,3(mol);n_{C_{12}H_{22}O_{11}}=0,2(mol)`

`C_{12}H_{22}O_{11}+H_2O` $\xrightarrow{t^o,H^+}$ `2C_6H_{12}O_6`

`C_5H_{11}O_5CHO` $\xrightarrow{AgNO_3//NH_3}$ `2Ag`

`->n_{Ag}=2n_{C_6H_{12}O_6\ spu}=2(0,2.2+0,3)=1,4(mol)`

`->m_{Ag}=1,4.108=151,2(g)`