`n_{Ag} = (90,72)/(108) = 0,84(mol)`

Đặt $n_{maltose} = 3a (mol); n_{saccharose} = 2a (mol)$

$n_{glucose} = 2n_{maltose(pư)} + n_{saccharose(pư)} = 2.3a.80\% + 2a.75\% = 6,3a(mol)$

$n_{maltose(dư)} = 3a - 3a.80\% = 0,6a(mol)$
$n_{fructose} = n_{saccharose(pư)} = 2a.75\% = 1,5a(mol)$
Ta có:

$n_{Ag} = 2n_{glucose} + 2n_{maltose} + 2n_{fructose}$
`=> 0,84 = 2.6,3a + 2.0,6a + 2.1,5a`

`=> a = 0,05(mol)`

`=> m = 0,05.(3 + 2).342 = 85,5(g)`