Câu 5

a)MgCO3 + 2HCl -> MgCl2 + H2O + CO2

b)nMgCO3=8.4/84=0.1mol

MgCO3 + 2HCl -> MgCl2 + H2O + CO2

(mol) 0.1 0.2 0.1 0.1

HCl = 0.2*36.5=7.3g

mdd= mMgCO3 + mddHCl -mCO2

=8.4+146-0.1*44=150g

C% HCl = 7.3/150*100=4.86%

c)mMgCl2=0.1*95=9.5g

C%MgCl2=9.5/150*100=6.33%

Câu 5

a) MgCO₃ + 2HCl \(\rightarrow\) MgCl₂ + H₂O + CO₂

b) n_MgCO3 = 8,4 : 84 = 0,1 (mol)

=> n_HCl = 2n_MgCO3 = 0,2 (mol)

=> m_HCl = 0,2 . 36,5 = 7,3 (mol)

C%(dd HCl) = \(\dfrac{ct}{dd}\) . 100% = \(\dfrac{7,3}{146}\) .100% = 5%

c) **Mình chưa hỉu đề bài**

Câu 6

a) CaCO₃ + 2HCl \(\rightarrow\) CaCl₂ + H₂O + CO₂

b) ...

*** Bạn giải thích lại hộ mk cái đề rùi mk giải nốt cho _ Hihi***

Caau 6

a) CaCO3 + 2HCl -> CaCl2 + H2O + CO2

b)nCaCO3 = 10/100=0.1mol

mHCl = 114.1*8/100=9.128g

nHCl = 9.128/36.5=0.25mol

CaCO3 + 2HCl -> CaCl2 + H2O + CO2

(mol) 1 2

(mol) 0.1 0.25

Lập tỉ lệ: 0.1/1<0.25/2. Vậy HCl dư

CaCO3 + 2HCl -> CaCl2 + H2O + CO2

(mol) 0.1 0.2 0.1 0.2

mdd=mCaCO3 + mddHCl - mCO2

=10+114.1-0.2*44=115.3g

nHCl dư = 0.25-0.2=0.5mol

mHCl dư = 0.25*22.4=5.6g

mCaCl2= 0.1*111=11.1g

C%HCl dư=5.6/115.3*100=4.86%

C%CaCl2=11.1/115.3*100=9.63%

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

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

 Xem hình!

\(CaCO_3+ HCl → CaCl_2+H_2O +CO_2\)

\(n_{CaCO_3}=\dfrac{10}{40+12+16.3}=0,1(mol)\)

\(n_{HCl}=\dfrac{146}{1+35,5}=4(mol)\)

\(\Rightarrow n_{HCl_{dư}}=4-0,1=3,9(mol) ; n_{CaCl_2}=0,1(mol)\\\Rightarrow m_{\text{chất tan}} = m_{HCl_{dư}}+m_{CaCl_2}\\=0,39.(35,5+1)+0,1(40+35,5.2)=25,335(g)\)

Vậy...

PT: \(CaCO_3+2HCl→CaCl_2+H_2O+CO_2\)

\(n_{HCl_{dư}} = 4-0,1.2=3,8(mol) \Rightarrow m_{HCl_{dư}}=138,7 (g) \\ n_{CaCl_2}=0,1(mol) \Rightarrow m_{CaCl_2}=11,1(g) \\ m_{\text{chất tan}}=149,8 (g) \\ \Rightarrow C\%=\dfrac{149,8}{146+10} .100\%=96\%\).

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

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

n_MgCO3 = \(\frac{m}{M}\)= \(\frac{8,4}{84}\)= 0,1 (mol)

Khi cho MgCO₃ vào HCl, ta có PTHH:

a. MgCO₃ + 2HCl \(\rightarrow\)MgCl₂ + CO₂\(\uparrow\)+ H₂O

        0,1 \(\rightarrow\)0,2      :           0,1  :    0,1     : 0,1            (mol)

b. C%_HCl = \(\frac{mt}{md}\). 100% = \(\frac{36,5.0,2}{146}\).100% = 5 %

c. m_ddsau = m_MgCO3 + m_HCl - m_CO2 = 8,4 + 146 - 44.0,1 = 150 (g)

C%_MgCl2 = \(\frac{mt}{md}\).100% = \(\frac{0,1.95}{150}\).100% \(\approx\) 6,33 %

a) PTHH :\(Mg+2HCl-->MgCl_2+H_2\)   (1)

                \(Fe+2HCl-->FeCl_2+H_2\)      (2)

b) \(n_{HCl}=\dfrac{600\cdot3,65}{100\cdot36,5}=0,6\left(mol\right)\)

Đặt \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\) => 24x + 56y = 13,6    (*)

Theo PTHH (1) và (2) : \(\Sigma n_{HCl}=2n_{Mg}+2n_{Fe}\)

=> \(0,6=2x+2y\)   (**)

Từ (*) và (**) suy ra : \(\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

=> \(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1\cdot24}{0,1\cdot24+0,2\cdot56}\cdot100\%\approx17,65\%\\\%m_{Fe}=100\%-17,65\%=82,35\%\end{matrix}\right.\)

c) Theo PTHH (1) và (2) :

\(n_{MgCl_2}=n_{Mg}=0,1\left(mol\right)\)

\(n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)

Theo ĐLBTKL :

\(m_A+m_{ddHCl}=m_{dd.spu}\)

=> 13,6 + 600 = m_(dd.spu)

=> m_(dd.spu) = 613,6 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,1\cdot95}{613,6}\cdot100\%\approx1,55\%\\C\%_{FeCl_2}=\dfrac{0,2\cdot127}{613,6}\cdot100\%\approx4,14\%\end{matrix}\right.\)

bạn sửa giùm mình phần c từ đoạn : trước khi áp dụng ĐLBTKL nha :

Theo PTHH (1) và (2) : \(n_{H2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\)

Theo ĐLBTKL :

m_A + m_(ddHCl) = m_(ddspu) + m_H2

=> 13,6 + 600 = m_(ddspu) + 2.0,3

=> m_(ddspu) = 613 (g)

=> \(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,1\cdot95}{613}\cdot100\%\approx1,55\%\\C\%_{FeCl_2}=\dfrac{0,2\cdot127}{613}\cdot100\%\approx4,14\%\end{matrix}\right.\)