Đáp án:

 3,36l

0,3l

360g

3,92%

Giải thích các bước giải:

\(\begin{array}{l}
Mg + 2HCl \to MgC{l_2} + {H_2}\\
{n_{Mg}} = \dfrac{{3,6}}{{24}} = 0,15\,mol\\
{n_{{H_2}}} = {n_{Mg}} = 0,15\,mol\\
{V_{{H_2}}} = 0,15 \times 22,4 = 3,36l\\
{n_{HCl}} = 2{n_{Mg}} = 0,3\,mol\\
{V_{HCl}} = \dfrac{{0,3}}{1} = 0,3l\\
{m_{{\rm{dd}}HCl}} = 300 \times 1,2 = 360g\\
{m_{{\rm{dd}}spu}} = 3,6 + 360 - 0,15 \times 2 = 363,3g\\
{C_\% }MgC{l_2} = \dfrac{{0,15 \times 95}}{{363,3}} \times 100\%  = 3,92\% 
\end{array}\)

`n_{Mg}=\frac{3,6}{24}=0,15(mol)`

`Mg+2HCl\to MgCl_2+H_2`

`n_{H_2}=n_{Mg}=0,15(mol)`

`=> V_{H_2}=0,15.22,4=3,36(l)`

`n_{HCl}=2n_{H_2}=0,3(mol)`

`V_{HCl}=\frac{0,3}{1}=0,3(l)`

`m_{\text{dd HCl}}=d.V=300.1,2=360g`

BTKL:

`m_{\text{dd spu}}=m_{Mg}+m_{\text{dd} HCl}-m_{H_2}`

`=> m_{\text{dd spu}}=3,6+360-0,15.2=363,3g`

`C%_{MgCl_2}=\frac{95.0,15.100%}{363,3}\approx 3,92%`