Câu 1 : 

\(n_{Mg}=\dfrac{8.4}{24}=0.35\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(0.35.......0.7.........0.35..........0.35\)

\(C\%_{HCl}=\dfrac{0.7\cdot36.5}{146}\cdot100\%=17.5\%\)

\(m_{\text{dung dịch sau phản ứng}}=8.4+146-0.35\cdot2=153.7\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{0.35\cdot95}{153.7}\cdot100\%=21.6\%\)

Câu 2 :

\(n_{CaCO_3}=\dfrac{10}{100}=0.1\left(mol\right)\)

\(n_{HCl}=\dfrac{114.1\cdot8\%}{36.5}=0.25\left(mol\right)\)

\(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

\(1................2\)

\(0.1.............0.25\)

\(LTL:\dfrac{0.1}{1}< \dfrac{0.25}{2}\Rightarrow HCldư\)

\(m_{\text{dung dịch sau phản ứng}}=10+114.1-0.1\cdot44=119.7\left(g\right)\)

\(C\%_{HCl\left(dư\right)}=\dfrac{\left(0.25-0.2\right)\cdot36.5}{119.7}\cdot100\%=1.52\%\)

\(C\%_{CaCl_2}=\dfrac{0.2\cdot111}{119.7}\cdot100\%=18.54\%\)

Câu 3 : 

\(n_{H_2}=\dfrac{6.72}{22.4}=0.3\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(0.3.......0.6......................0.3\)

\(m_{Fe}=0.3\cdot56=16.8\left(g\right)\)

\(m_{dd_{HCl}}=150\cdot1.2=180\left(g\right)\)

\(C\%_{HCl}=\dfrac{0.6\cdot36.5}{180}\cdot100\%=12.16\%\)

\(C_{M_{HCl}}=\dfrac{0.6}{0.15}=4\left(M\right)\)

a)

$NaOH  + HCl \to NaCl + H_2O$
$OH^- + H^+ \to H_2O$

b)

$n_{HCl} = 0,1.0,01 = 0,001(mol)$
$n_{NaOH} = 0,2.0,5= 0,1(mol)$

$\Rightarrow$ NaOH dư, HCl hết

$n_{NaOH\ pư} = 0,001 \Rightarrow n_{NaOH\ dư} = 0,1 - 0,001 = 0,099(mol)$

$\Rightarrow [OH^-] = \dfrac{0,099}{0,1 + 0,2} = 0,33M$
$\Rightarrow pOH = -log(0,33) = 0,48 $

$pH = 14 - pOH = 14 - 0,48 = 13,52$

NaF + HCl → NaCl + HF

H+ + F+ → HF

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

Ta có: \(\left\{{}\begin{matrix}n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\\n_{HCl}=\dfrac{109,5\cdot10\%}{36,5}=0,3\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\) \(\Rightarrow\) HCl còn dư, Magie p/ứ hết

\(\Rightarrow n_{Mg}=n_{MgCl_2}=n_{H_2}=0,1\left(mol\right)=n_{HCl\left(dư\right)}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\\m_{HCl\left(dư\right)}=0,1\cdot36,5=3,65\left(g\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(saup/ứ\right)}=m_{Mg}+m_{ddHCl}-m_{H_2}=111,7\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{9,5}{111,7}\cdot100\%\approx8,5\%\\C\%_{HCl\left(dư\right)}=\dfrac{3,65}{111,7}\cdot100\%\approx3,27\%\end{matrix}\right.\)

NaHCO3 + HCl → NaCl + H2O + CO2↑

H+ + HCO3- → H2O + CO2↑

Đáp án C. Vì chỉ rõ các ion nào đã tác dụng với nhau làm cho phản ứng xảy ra.