Ta có
V = $\dfrac{100}{2}$ = 50 (ml)
$n_{Pb(NO_3)_2}$ = 0,025(mol)
$n_{AgNO_3}$ = 0,03(mol)
$n_{Al}$ = $\dfrac{1,2}{27}$ =$ \dfrac{2}{45}$(mol)
$2Al+3Pb(NO_3)_2--->2Al(NO_3)_3+3Pb$
$Al+3AgNO_3--->Al(NO3)_3+3Ag$
$n_{Al dư}$ =$ \dfrac{2}{45}$ - $\dfrac{0,025.2}{3}$ - $\dfrac{0,03}{3}$ =...