Đáp án:

\({V_{{H_2}}} = 1,12{\text{ lít}}\)

Giải thích các bước giải:

 Phản ứng xảy ra:

\(Mg + 2HCl\xrightarrow{{}}MgC{l_2} + {H_2}\)

Ta có:

\({n_{Mg}} = \frac{{1,2}}{{24}} = 0,05{\text{ mol}}\)

\({m_{HCl}} = 54,75.10\%  = 5,475{\text{ gam}}\)

\( \to {n_{HCl}} = \frac{{5,475}}{{36,5}} = 0,15{\text{ mol > 2}}{{\text{n}}_{Mg}}\)

Vậy \(HCl\) dư.

\( \to {n_{{H_2}}} = {n_{Mg}} = 0,05{\text{ mol}}\)

\( \to {V_{{H_2}}} = 0,05.22,4 = 1,12{\text{ lít}}\)

BTKL: 

\({m_{dd{\text{ sau phản ứng}}}} = {m_{Mg}} + {m_{dd{\text{ HCl}}}} - {m_{{H_2}}}\)

\( = 1,2 + 54,75 - 0,05.2 = 55,85{\text{ gam}}\)

\({n_{MgC{l_2}}} = {n_{Mg}} = 0,05{\text{ mol}}\)

\({n_{HCl{\text{ dư}}}} = 0,15 - 0,05.2 = 0,05{\text{mol}}\)

\( \to C{\% _{MgC{l_2}}} = \frac{{0,05.(24 + 35,5.2)}}{{55,85}} = 8,5\% \)

\(C{\% _{HCl{\text{ dư}}}} = \frac{{0,05.36,5}}{{55,85}} = 3,267\% \)

Đáp án:

\(\begin{array}{l} b,\ V_{H_2}=1,12\ lít.\\ c,\ C\%_{MgCl_2}=8,5\%\\ C\%_{HCl}\text{(dư)}=3,27\%\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\ PTHH:Mg+2HCl\to MgCl_2+H_2↑\\ b,\ n_{Mg}=\dfrac{1,2}{24}=0,05\ mol.\\ n_{HCl}=\dfrac{54,75\times 10\%}{36,5}=0,15\ mol.\\ \text{Lập tỉ lệ:}\ \dfrac{0,05}{1}<\dfrac{0,15}{2}\\ \Rightarrow HCl\ \text{dư.}\\ Theo\ pt:\ n_{H_2}=n_{Mg}=0,05\ mol.\\ \Rightarrow V_{H_2}=0,05\times 22,4=1,12\ lít.\\ c,\ Theo\ pt:\ n_{MgCl_2}=n_{Mg}=0,05\ mol.\\ m_{\text{dd spư}}=m_{Mg}+m_{\text{dd HCl}}-m_{H_2}\\ \Rightarrow m_{\text{dd spư}}=1,2+54,75-0,05\times 2=55,85\ g.\\ n_{HCl}\text{(dư)}=0,15-(0,05\times 2)=0,05\ mol.\\ \Rightarrow C\%_{MgCl_2}=\dfrac{0,05\times 95}{55,85}\times 100\%=8,5\%\\ C\%_{HCl}\text{(dư)}=\dfrac{0,05\times 36,5}{55,85}\times 100\%=3,27\%\end{array}\)

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