18)

Ta có:

\({C_{M{\text{ MgC}}{{\text{l}}_2}}} = \frac{{{n_{MgC{l_2}}}}}{{{V_{dd}}}} = \frac{{0,2}}{2} = 0,1{\text{ M}}\)

1)

Phản ứng xảy ra:

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

Ta có:

\({n_{Mg}} = \frac{{2,4}}{{24}} = 0,1{\text{ mol}}\)

\({n_{HCl}} = 100.14,6\%  = 14,6{\text{ gam}} \to {{\text{n}}_{HCl}} = \frac{{14,6}}{{36,5}} = 0,4{\text{ mol}}\)

\( \to {n_{HCl}} > 2{n_{Mg}}\) nên \(HCl\) dư

\( \to {n_{MgC{l_2}}} = {n_{{H_2}}} = {n_{Mg}} = 0,1{\text{ mol}}\)

\( \to {V_{{H_2}}} = 0,1.22,4 = 2,24{\text{ lít}}\)

BTKL:

\({m_{Mg}} + {m_{dd\;{\text{HCl}}}} = {m_{dd}} + {m_{{H_2}}}\)

\( \to 2,4 + 100 = {m_{dd}} + 0,1.2 \to {m_{{\text{d}}d}} = 102,2{\text{ gam}}\)

Ta có:

\({m_{MgC{l_2}}} = 0,1.95 = 9,5{\text{ gam}}\)

\({n_{HCl{\text{ dư}}}} = 0,4 - 0,1.2 = 0,2{\text{ mol}}\)

\( \to {m_{HCl{\text{ dư}}}} = 0,2.36,5 = 7,3{\text{ gam}}\)

\( \to C{\% _{MgC{l_2}}} = \frac{{9,5}}{{102,2}} = 9,3\% \)

\(C{\% _{HCl{\text{ dư}}}} = \frac{{7,3}}{{102,2}} = 7,14\% \)

Đáp án:

 

Giải thích các bước giải:

 18)0,1M

II)

Câu-1Phản ứng xảy ra:

Mg+2HCl---->MgCl2+H2

Ta-sẽ-có:

nMg=2,4/24=0,1mol

nHCl=100.14,6=14.6g---->nhcL=14,6/36,5=0,4mol

nHCl>2nMg-nênHCl-dư

nMgCl2=nH2=nMg=0,1mol

VH2=0,1.22,4=2,24l

mMg+mdd=mdd+mH2

2,4+100=mdd+0,1.2--->mdd=102,2g

Ta-có

mMgCl2=0,1.95=9,5g

mHCl-dư=0,4-0,1.2=0,2mol--->mHCl-dư=0,2.36,5=7,3g

C%MgCl2=9,5/102,2=9,3%

C%Hcl=7,3/102,2=7,14%

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