a) PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

b) Ta có: \(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\) 

\(\Rightarrow\left\{{}\begin{matrix}n_{MgCl_2}=0,2\left(mol\right)\\n_{HCl}=0,4\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{MgCl_2}=0,2\cdot95=19\left(g\right)\\C\%_{HCl}=\dfrac{0,4\cdot36,5}{200}\cdot100\%=7,3\%\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{MgO}+m_{ddHCl}=208\left(g\right)\)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{19}{208}\cdot100\%\approx9,13\%\)

c) PTHH: \(MgCl_2+2NaOH\rightarrow2NaCl+Mg\left(OH\right)_2\downarrow\)

Ta có: \(\left\{{}\begin{matrix}n_{MgCl_2}=0,2\left(mol\right)\\n_{NaOH}=\dfrac{200\cdot4\%}{40}=0,2\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\) \(\Rightarrow\) NaOH p/ứ hết, MgCl₂ còn dư

\(\Rightarrow\left\{{}\begin{matrix}n_{NaCl}=0,2\left(mol\right)\\n_{Mg\left(OH\right)_2}=0,1\left(mol\right)=n_{MgCl_2\left(dư\right)}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\\m_{MgCl_2\left(dư\right)}=9,5\left(g\right)\\m_{Mg\left(OH\right)_2}=0,1\cdot58=5,8\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{ddA}+m_{ddNaOH}-m_{Mg\left(OH\right)_2}=402,2\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{11,7}{402,2}\cdot100\%\approx2,91\%\\C\%_{MgCl_2\left(dư\right)}=\dfrac{9,5}{402,2}\cdot100\%\approx2,36\%\end{matrix}\right.\) 

Theo gt ta có: $n_{MgO}=0,2(mol)$

a, $MgO+2HCl\rightarrow MgCl_2+H_2O$

b, Ta có: $n_{HCl}=0,4(mol)\Rightarrow x=7,3$

Bảo toàn khối lượng ta có: $m_{ddA}=208(g)$

$\Rightarrow \%C_{MgCl_2}=9,13\%$

c, Ta có: $n_{NaOH}=0,2(mol)$

$\Rightarrow n_{Mg(OH)_2}=0,1(mol)$

Bảo toàn khối lượng ta có: $m_{ddB}=208+200-0,1.58=402,2(g)$

$\Rightarrow \%C_{MgCl_2}=2,36\%$

a)

$MgO + 2HCl \to MgCl_2  + H_2O$

b)

$n_{MgO} = \dfrac{8}{40} = 0,2(mol)$

$n_{HCl} = 2n_{MgO} = 0,4(mol)$

$C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%$

(x = 7,3)

$m_{dd\ A} = 8 + 200 = 208(gam)$
$C\%_{MgCl_2} = \dfrac{0,2.95}{208}.100\% = 9,13\%$

\(m_{NaOH}=\dfrac{200\cdot8}{100}=16\left(g\right)\Rightarrow n_{NaOH}=\dfrac{16}{40}=0,4mol\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

 0,4           0,4         0,4        0,4

a)\(m_{HCl}=0,4\cdot36,5=14,6\left(g\right)\) 

 \(\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3}\cdot100=200\left(g\right)\)

b)\(m_{NaCl}=0,4\cdot58,5=23,4\left(g\right)\)

   \(m_{H_2O}=0,4\cdot18=7,2\left(g\right)\)

   \(m_{ddsau}=200+200-7,2=392,8\left(g\right)\)

   \(\Rightarrow C\%=\dfrac{23,4}{392,8}\cdot100=5,96\%\)

c) \(n_{SO_2}=\dfrac{6,72}{22,4}=0,3mol\)

     \(2NaOH+SO_2\rightarrow Na_2SO_4+H_2O\)

         0,4          0,3         0,3              0,3

     \(m_{Na_2SO_4}=0,3\cdot142=42,6\left(g\right)\)

Ta có pthh

Na2O + H2O2NaOH

Áp dụng định luật bảo toàn khối lượng ta có

mNa2O+mH2O=mNaOH

mNaOH=6,2+93,8=100 g

Ta có

nH2O=6,262=0,1mol6,262=0,1���

Theo pthh

nNa2O=

Vậy nồng độ của dd tạo thành là 8%

\(a,n_{Na_2O}=\dfrac{6,2}{62}=0,1mol\\ Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=0,1.2=0,2mol\\ C_{\%A}=C_{\%NaOH}=\dfrac{0,2.40}{6,2+93,8}\cdot100\%=4\%\\ b.n_{CuSO_4}=\dfrac{200.16\%}{100\%.160}=0,2mol\\ CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ \Rightarrow\dfrac{0,2}{1}>\dfrac{0,2}{2}\Rightarrow CuSO_4.dư\\ n_{Na_2SO_4}=n_{Cu\left(OH\right)_2}=n_{CuSO_4,pư}=0,2:2=0,1mol\\ m_{ddA}=6,2+93,8+200-0,1.98=290,2g\\ C_{\%Na_2SO_4}=\dfrac{0,1.142}{290,2}\cdot100\%\approx4,98\%\\ C_{\%CuSO_4,dư}=\dfrac{\left(0,2-0,1\right).160}{290,2}\cdot100\%\approx5,51\%\)