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

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

Zn + 2HCl --> ZnCl2 + H2

0,5      1              0,5       0,5

nZn=\(\dfrac{32,5}{65}\)=0,5(mol)

mZnCl2=0,5.136=68(g)

VH2=0,5.22,4=11,2(l)

c) CM HCl=\(\dfrac{1}{0,25}=4M\)

Giúp với mng ơi

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

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

\(0.2...................0.2..........0.2\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{FeCl_2}=0.2\cdot127=25.4\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=11.2+200-0.2\cdot2=210.8\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{25.4}{210.8}\cdot100\%=12.05\%\)

Giúp với mai mình thi rồi!

ta có lượng \(H^+\) có trong dung dịch là :

\(n_{H^+}=2n_{H_2SO_4}+n_{HCL}=2\times0,2\times1+0,2\times2=0,8\left(mol\right)\)

a. ta có \(n_{H_2}=\frac{1}{2}n_{H^+}=0,4mol\Rightarrow V_{H_2}=22,4\times0,4=8,96\left(lit\right)\)

b. ta có \(m_{\text{hỗn hợp}}+m_{\text{axit }}=m_{\text{chất tan}}+m_{\text{ khí}}\)

nên \(m_{\text{chất tan }}=12,9+0,2\times98+0,4\times36,5-0,4\times2=46,3\left(g\right)\)

.a)a)

PTHH:Fe+2HCl−−−>FeCl2+H2(1)

Fe2O3+6HCl−−−>2FeCl3+3H2O(2)

nH2=2,2422,4=0,1(mol)n

Theo (1) nFe=nH2=0,1(mol)

=>mFe=0,1.56=5,6(g)

b)

mFe2O3 = 8 - 5,6 = 2,4 (g)

=> nFe2O3 = 0,015 (mol)

Theo (1) và (2), nHCl = nHCl(1) + nHCl(2)

<=> nHCl = 0,2 + 0,09 = 0,29 (mol)

=> mHCl = 0,29. 36,5 = 10,585 (g)

Ta có: mddHCl = mHCl.100= 105,85 (g)

=> VddHCl= mddHCl\DHCl= 105,85\1,1= 96,23 (ml)

Đổi 96,23 ml = 0,09623 lit

Theo (1) nFeCl2 = 0,1 (mol)

=> CMFeCl2 = 1,04 M

Theo (2) nFeCl3 = 2.nFe2O3 = 2. 0,015 = 0,03 (mol)

=>CMFeCl3 = 0,3 M

3>

a)

Zn+2HCl->ZnCl2+H2

ZnO+2HCl->ZnCl2+H2O

nH2=3,36/22,4=0,15 mol

=>nZn=0,15 mol

mZn=0,15 x65=9,75 g

mZnO=17,85-9,75=8,1 g

b)

nZnO=8,1/81=0,1 mol

=>nHCl=0,1x2+0,15x2=0,5 mol

VHCl=0,5/1=0,5 l

nFe = 5.6/56 = 0.1 (mol) 

nHCl = 0.2*2 = 0.4 (mol) 

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

LTL : 0.1/1 < 0.4/2 => HCl dư 

mHCl dư = ( 0.4 - 0.2 ) * 36.5 = 7.3 (g) 

VH2 = 0.2*22.4 = 4.48 (l) 

CM FeCl2 = 0.1/0.2 = 0.5(M)

CM HCl dư = 0.2 / 0.2 = 1(M)