a) $n_{Al_2(SO_4)_3}=\dfrac{75,24}{2×27+3×(32+4×16)}≈0,22(mol)$
b) $n_{O_2}=\dfrac{15,68}{22,4}=0,7(mol)$
c) $n_{H_2SO_4}=\dfrac{13,2×10^{23}}{6×10^{23}}=2,2(mol)$
d) $n_{Fe}=\dfrac{11,2}{56}=0,2(mol)$
$n_{Al}=\dfrac{3,24}{27}=0,12(mol)$
⇒ $n_{X}=n_{Fe}+n_{Al}=0,2+0,12=0,32(mol)$
e) $n_{O_2}=\dfrac{8,94}{22,4}≈0,4(mol)$
$n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)$
⇒ $n_{Y}=n_{O_2}+n_{H_2}=0,4+0,1=0,5(mol)$