Đáp án:

CM_{HCl}=5M

Giải thích các bước giải:

$$4Al + 3O_2 \xrightarrow{t^0} 2Al_2O_3$$

$$2Cu + O_2 \xrightarrow{t^0} 2CuO$$

$$2Zn + O_2 \xrightarrow{t^0} 2ZnO$$

Al_2O_3 + 6HCl \to 2AlCl_3 + 3H_2O

CuO + 2HCl \to CuCl_2 + H_2O

ZnO + 2HCl \to ZnCl_2 + H_2O

BTKL: m_{O_2}=52,6-36,6=16g

=> n_{O_2}=16:32=0,5 mol

Theo pt: $$n_{HCl}=2.n_{H_2O}=2.n_{O(Oxit)}=2.2.n_{O_2}=4.0,5=2 mol$$

=> CM_{HCl}={2}/{0,4}=5M

$$m_{O_{2}}=m_{oxit}-m_{Kim~loại}=52,6-36,6=16~(mol)$$.

$$n_{O_{2}}=$$ $$\frac{m}{32}=$$ $$\frac{16}{32}=0,5~(~mol~)$$

$$∑$$$$n_{O}~trong~O_{2}=2~.~0,5=1~(~mol~)$$

.

$$∑n_{O}~trong~O_{2}=∑n_{O}~trong~H_{2}O=~1~(~mol~)$$

$$-->n_{H_{2}O}=~1~(~mol~)$$

    $$-->n_{HCl}=2~.~n_{H_{2}O}=2~.~1=2~(~mol~)$$

$$C_{M_{(HCl)}}=$$$$\frac{n_{chất~tan}}{V_{dung~dịch}}=$$ $$\frac{2}{0,4}=5~M$$