1. С3Н7ОН+HCl=C3H7Cl+H2O
2.2C3H7Cl+Zn=C6H14+ZnCl2
3.C6H14⇒C6H6+4H2↑
4.C6H6+HNO3(H2SO4)⇒C6H5NO2+H2O
5.C6H5NO2+6H⇒C6H5NH2+2H2O
6.2C6H5NH2+9,5O2⇒12CO2+7H2O+N2↑
2Al + 6HCl = 2AlCl3 + 3H2
n (H2) = 5,6 / 22,4 = 0,25 моль
n(H2) : n(Al) = 3:2
n(Al) = 0,166 моль
m (Al) = 0,166 x 27 = 4,482 г
Ответ: 4,482г
14,88 кг меди может быть получено
<span><span>Fe3O4</span> → <span>3Fe</span>+<span>2<span>O2
</span></span></span><span><span>Fe</span>+<span>H2S</span> → <span>FeS</span>+<span>H<span>2
X = Fe
</span></span></span>