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

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

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 : 

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

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

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

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

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

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

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

\(CaCO_3+ HCl → CaCl_2+H_2O +CO_2\)

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

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

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

Vậy...

PT: \(CaCO_3+2HCl→CaCl_2+H_2O+CO_2\)

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

a, PT: \(CaCO_3+2HCl\rightarrow CaCl_2+CO_2+H_2O\)

\(MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\)

Ta có: 100n_CaCO3 + 84n_MgCO3 = 14,2 (1)

Theo PT: \(n_{CO_2}=n_{CaCO_3}+n_{MgCO_3}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{CaCO_3}=0,1\left(mol\right)\\n_{MgCO_3}=0,05\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CaCO_3}=\dfrac{0,1.100}{14,2}.100\%\approx70,42\%\\\%m_{MgCO_3}\approx29,58\%\end{matrix}\right.\)

b, Theo PT: \(n_{HCl}=2n_{CO_2}=0,3\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,6}=0,5\left(M\right)\)

PTHH :

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

x                 2x               x           x          x 

\(MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2\uparrow\)

y                 2y               y              y          y 

Có:

\(\left\{{}\begin{matrix}100x+84y=14,2\\x+y=\dfrac{3,36}{22,4}=0,15\end{matrix}\right.\)

\(\Rightarrow x=0,1;y=0,05\)

\(a,\%m_{CaCO_3}=0,1.100:14,2.100\%\approx72,423\%\)

\(\%m_{MgCO_3}=100\%-72,423\%\approx29,577\%\)

\(b,C_{M\left(HCl\right)}=\dfrac{0,2+0,1}{0,6}=0,5\left(M\right)\)

Gọi số mol của CaCO3 và MgCO3 lần lượt là x và y => mhỗn hợp = PT(1)

Phương trình hóa học:

CaCO3 to→→�� CaO + CO2

  x mol            →      x mol

MgCO3 to→→�� MgO + CO2

  y mol           →         y mol

=>nCO2=PT(2)

nCaCO3=10100=0,1molnCaCO3=10100=0,1mol, nHCl=mdd.C%36,5.100=114,1.836,5.100=0,25molnHCl=mdd.C%36,5.100=114,1.836,5.100=0,25mol

CaCO3 + 2HCl  →CaCl2+ CO2↑ + H2O

Ta có: 0,11<0,2520,11<0,252 → CaCO3 hết

Theo PTHH: nCO2=nCaCl2= nCaCO3= 0,1 mol

Khối lượng dung dịch sau phản ứng là: mdd= mCaCO3+ mdd HCl- mCO2= 10 + 114,1- 0,1.44=119,7 gam

→C%CaCl2=mCaCl2.100%mdd=0,1.111.100%119,7=9,27%→C%CaCl2=mCaCl2.100%mdd=0,1.111.100%119,7=9,27%

CaCO₃ + 2HCl ----------> CaCl₂ + H₂O + CO₂ (1)

0,1 0,2 0,1 (mol)

n_CaCO3 = 10 /100 =0,1 (mol)

m_HCl = 114,1 . 8% : 100% = 9,128 (g)

=> n _HCl =9,128 /26,5 = 0,25 (mol)

n_HCl = 10/100 = 0,1 (mol)

nCaCO3 /1 = 0,1/1 = 0,1 < nHCl /2 = 0,25/2 =0,125

=> HCl dư

=> sau pư thu được dd CaCl₂ , dd HCl dư

C% _{dd ZnCl2} = \(\frac{0,1.136}{10+114,1}.100\%\) = 10,96%

C% _{dd HCl dư} = \(\frac{\left(025-0,2\right).36,5}{10+114,1}.100\%\)= 1,47%

PTHH: \(CaO+2HCl\rightarrow CaCl_2+H_2O\)  (1)

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

a) Ta có: \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)=n_{CaCO_3}\)

\(\Rightarrow m_{CaCO_3}=0,2\cdot100=20\left(g\right)\) \(\Rightarrow\%m_{CaCO_3}=\dfrac{20}{25,6}\cdot100\%=78,125\%\)

\(\Rightarrow\%m_{CaO}=21,875\%\)

b) Theo 2 PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(2\right)}=2n_{CaCO_3}=0,4mol\\n_{HCl\left(1\right)}=2n_{CaO}=2\cdot\dfrac{25,6-20}{56}=0,2mol\end{matrix}\right.\)

\(\Rightarrow\Sigma n_{HCl}=0,6mol\) \(\Rightarrow C\%_{HCl}=\dfrac{0,6\cdot36,5}{210\cdot1,05}\cdot100\%\approx9,93\%\)