Bạn tham khảo:
$Fe+4HNO_3 \to Fe(NO_3)_3+NO+2H_2O$
$Al+4HNO_3 \to Al(NO_3)_3+NO+2H_2O$
$n_{Fe}=a(mol)$
$n_{Al}=b(mol)$
$m_{hh}=56a+27b=11(g)(1)$
$n_{NO}=a+b=\frac{6,72}{22,4}=0,3(mol)(2)$
$a=0,1; b=0,2$
$a/$
$m_{Fe}=0,1.56=5,6(g)$
$m_{Al}=5,4(g)$
$b/$
$3KOH+Fe(NO_3)_3 \to Fe(OH)_3+KNO_3(1)$
$3KOH+Al(NO_3)_3 \to Al(OH)_3+KNO_3(1)$
$(1)$
$n_{KOH}=0,3(mol)$
$(2)$
$n_{KOH}=0,35-0,3=0,05(mol)$
$n_{Al(OH)_3}=\frac{1}{60}(mol)$
$m=0,1.107+\frac{1}{60}.78=12(g)$