$n_{Ag}= \dfrac{10,8}{108} = 0,1(mol)$
$C_6H_{12}O_6 + Ag_2O \xrightarrow{NH_3} 2Ag + C_6H_{12}O_7$
Theo PTHH :
$n_{glucozo\ pư} = \dfrac{1}{2}n_{Ag} = 0,05(mol)$
$\Rightarrow n_{glucozo\ ban\ đầu} = 0,05 : 75\% = \dfrac{1}{15}(mol)$
$\Rightarrow m = \dfrac{1}{15}.180 = 12(gam)$