Em tham khảo nha :

\(\begin{array}{l}
a)\\
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = \dfrac{{7,2}}{{24}} = 0,3mol\\
{n_{{H_2}}} = {n_{Mg}} = 0,3mol\\
{V_{{H_2}}} = 0,3 \times 24,79 = 7,437l\\
b)\\
{n_{HCl}} = 2{n_{Mg}} = 0,6mol\\
{C_{{M_{HCl}}}} = \dfrac{{0,6}}{{0,3}} = 2M\\
c)\\
{H_2} + CuO \to Cu + {H_2}O\\
{n_{Cu}} = {n_{{H_2}}} = 0,3mol\\
{m_{Cu}} = 0,3 \times 64 = 19,2g
\end{array}\)

Đáp án:

$a) V_{H_2} = 6,72(l)$

$b) C_{M(HCl} = 2M$

$c) m_{Cu} = 19,2 (g)$

Giải thích các bước giải:

$a) n_{Mg} = \dfrac{7,2}{24} = 0,3 (mol)$

PTHH: $Mg + 2HCl \rightarrow MgCl_2 + H_2(1)$

Theo PTHH $(1): n_{H_2} = n_{Mg} = 0,3 (mol)$

$\rightarrow V_{H_2} = 0,3.22,4 = 6,72 (l)$

$b)$ Theo PTHH $(1): n_{HCl} = 2n_{Mg} = 0,6 (mol)$

$\rightarrow C_{M(HCl} = \dfrac{0,6}{0,3} = 2M$

$c) $ PTHH: $CuO + H_2 \xrightarrow{t^o} Cu + H_2O(2)$

Theo PTHH $(2): n_{Cu} = n_{H_2} = 0,3 (mol)$

$\rightarrow m_{Cu} = 64.0,3 = 19,2 (g)$

$Kudo/Shinichi$