Mg+2HCl->MgCl2+H2

0,15--0,3--------------0,15

CuO+2HCl->CuCl2+H2O

0,1------0,2

n H2=\(\dfrac{3,36}{22,4}\)=0,15 mol
=>%m Mg=\(\dfrac{0,15.24}{11,6}.100=31,03\%\)

=>m CuO=8g =>n CuO=\(\dfrac{8}{80}\)=0,1 mol

=>%m CuO=68,97%

=>CM HCl=\(\dfrac{0,3+0,2}{0,2}\)=2,5M

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

b, Theo PTHH : \(n_{HCl}=2n_{MgO}=2.\dfrac{m}{M}=0,4\left(mol\right)\)

\(\Rightarrow x=7,3\%\)

Theo PTHH : \(n_{MgCl2}=n_{MgO}=0,2\left(mol\right)\)

\(\Rightarrow m_{MgCl2}=19\left(g\right)\)

Mà mdd = \(m_{MgO}+m_{ddHCl}=208\left(g\right)\)

\(\Rightarrow C\%_{MgCl2}=\dfrac{m}{m_{dd}}.100\%=9,13\%\)

c, \(PTHH:MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2+2NaCl\)

....................0,1..............0,2...............0,1.............0,2.......

Ta có : \(n_{NaOH}=0,2\left(mol\right)\)

=> mdd = \(m_{MgCl2}+m_{NaOH}-m_{Mg\left(OH\right)2}=213,2g\)

- Thấy sau phản ứng dung dịch B gồm NaCl ( 0,2 mol ), MgCl2 dư ( 0,1mol )

\(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=11,7g\\m_{MgCl2}=9,5g\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=5,5\%\\C\%_{MgCl2}=4,46\%\end{matrix}\right.\)

a) PT phân tử: \(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)

    PT ion: \(CaCO_3+2H^+\rightarrow Ca^{2+}+H_2O+CO_2\uparrow\)

b) Ta có: \(\left\{{}\begin{matrix}n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\\n_{HCl}=\dfrac{43,8\cdot20\%}{36,5}=0,24\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,24}{2}\) \(\Rightarrow\) HCl dư, tính theo CaCO₃

\(\Rightarrow\left\{{}\begin{matrix}n_{CaCl_2}=0,1\left(mol\right)=n_{CO_2}\\n_{HCl\left(dư\right)}=0,04\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaCl_2}=0,1\cdot111=11,1\left(g\right)\\m_{CO_2}=0,1\cdot44=4,4\left(g\right)\\m_{HCl\left(dư\right)}=0,04\cdot36,5=1,46\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{CaCO_3}+m_{ddHCl}-m_{CO_2}=49,4\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{11,1}{49,4}\cdot100\%\approx22,47\%\\C\%_{HCl\left(dư\right)}=\dfrac{1,46}{49,4}\cdot100\%\approx2,96\%\end{matrix}\right.\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ a,Zn+2HCl\to ZnCl_2+H_2\\ b,n_{ZnCl_2}=0,1(mol)\\ \Rightarrow m_{ZnCl_2}=0,1.136=13,6(g)\\ c,n_{Zn}=0,1(mol)\\ \Rightarrow \%_{Zn}=\dfrac{0,1.65}{20}.100\%=32,5\%\\ \Rightarrow \%_{Ag}=100\%-32,5\%=67,5\%\)

áp dụng công thức là ra

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)

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

CaO+2HCl->CaCl2+H2O

CaCO3+2HCl->CaCl2+H2O+CO2

nHCl=0,3 mol

ta có :56x+100y=11,7

2x+2y=0,3 mol

=>x=y=0,075

=>mCaCl2=0,15.111=16,65g

 Xem hình!

a. \(Fe+2HCl\rightarrow FeCl_2+H_2\)

b. Giải bài toán bằng cách lập PTHH.

c. 

\(n_{Fe}=\dfrac{11,2}{56}=0,2mol\\ n_{HCl}=0,2.0,25=0,05mol\\ \Rightarrow\dfrac{0,2}{1}>\dfrac{0,05}{2}\Rightarrow Fe.dư\)

Chất còn lại trong dung dịch sau phản ứng là \(FeCl_2\)

\(d.n_{FeCl_2}=0,05:2=0,025mol\\ C_{M_{FeCl_2}}=\dfrac{0,025}{0,25}=0,1M\)

\(n_{Zn}=\dfrac{6.5}{65}=0.1\left(mol\right)\)

\(n_{HCl}=\dfrac{73\cdot36.5\%}{36.5}=0.73\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

\(0.1.........0.73\)

\(LTL:\dfrac{0.1}{1}< \dfrac{0.73}{2}\rightarrow HCldư\)

\(V_{H_2}=0.1\cdot22.4=2.24\left(l\right)\)

\(m_{ZnCl_2}=0.1\cdot136=13.6\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=6.5+73-0.1\cdot2=79.3\left(g\right)\)

\(C\%_{ZnCl_2}=\dfrac{13.6}{79.3}\cdot100\%=17.15\%\)

\(C\%_{HCl\left(dư\right)}=\dfrac{\left(0.73-0.2\right)\cdot36.5}{79.3}\cdot100\%=25.4\%\)

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