n_{Ag}=0,8(mol);n_{Br_2}=0,22(mol)

P2:

C_5H_{11}O_5CHO+Br_2+H_2O->C_5H_{11}O_5COOH+2HBr

->n_{glucose}=n_{Br_2}=0,22(mol)

P1:

C_6H_{12}O_6+Ag_2O  $$\xrightarrow{NH_3}$$  2Ag+C_6H_{12}O_7

->n_{f r u c t o s e}+n_{glucose}=1/2n_{Ag}=0,4(mol)

->n_{f r u c t o s e}=0,4-0,22=0,18(mol)

Do cùng CTPT.

->\%m_{f r u c t o s e}={0,18}/{0,18+0,22}.100\%=45\%