Решение представлено во вложении
Al + Cl3 = AlCl3
<span>4HCl (конц.) + MnO</span>2<span> = MnCl</span>2<span> + 2H</span>2<span>O + Cl</span>2 <span>↑</span>
<span><span>2CuSO</span>4<span> + 2NaCl + SO</span>2<span> + 2H</span>2<span>O = 2CuCl ↓ + 2H</span>2<span>SO</span>4<span> + Na</span>2<span>SO</span>4</span>
<span><span>4HCl (конц.) + 2Cu + O</span>2<span> = 2CuCl</span>2<span> + 2H</span>2<span>O</span></span>