a) Ta có: \(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)

Gọi a và b lần lượt là số mol của Al và Zn

Bảo toàn mol e: \(3a+2b=1,4\)

                   Mà \(27a+65b=31,4\)

\(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2\cdot27}{31,4}\cdot100\%\approx17,2\%\\\%m_{Zn}=82,8\%\end{matrix}\right.\)

b) Bảo toàn nguyên tố: \(n_{HCl}=2n_{H_2}=1,4mol\)

\(\Rightarrow V_{HCl}=\dfrac{1,4}{2}=0,7\left(l\right)=700\left(ml\right)\)

Đặt : 

nAl = a mol 

nZn = b mol 

mB = 27a + 65b = 31.4 (g) (1) 

2Al + 6HCl => 2AlCl3 + 3H2 

a___________________1.5a

Zn + 2HCl => ZnCl2 + H2 

b__________________b

nH2 = 1.5a + b = 15.68/22.4 = 0.7 (mol) (2) 

(1) , (2) : 

a = 0.2

b = 0.4 

%Al = 5.4/31.4 * 100% = 17.19%

%Zn = 100 - 17.19 = 82.81%

nHCl = 2nH2 = 0.7*2 = 1.4 (mol) 

Vdd HCl = 1.4 / 2 = 0.7 (l) 

PTHH: \(2KMnO_4+16HCl_{\left(đ\right)}\rightarrow2KCl+2MnCl_2+5Cl_2\uparrow+8H_2O\)

Ta có: \(n_{KMnO_4}=\dfrac{14,2}{158}=\dfrac{71}{790}\left(mol\right)\) 

\(\Rightarrow n_{Cl_2}=\dfrac{71}{316}\left(mol\right)\) \(\Rightarrow V_{Cl_2}=\dfrac{71}{316}\cdot22,4\approx5,03\left(l\right)\)

\(2A+2aHCl\rightarrow2ACl_a+aH_2\)

\(TheoPTHH:n_{H_2}=\dfrac{an_A}{2}=0,5an_A=0,5.a.\dfrac{1,17}{A}=\dfrac{V}{22,4}=0,015\)

\(\Rightarrow\dfrac{a}{A}=\dfrac{1}{39}\)

\(\Rightarrow\) A là Kali ( K ).

\(\Rightarrow n_{HCl}=2n_{H_2}=0,03\left(mol\right)\)

\(\Rightarrow V_{HCl}=\dfrac{n}{C_M}=\dfrac{0,03}{1,2}=0,025\left(l\right)=25\left(ml\right)\)

nHCl = 0.4 * 2.75 = 1.1 (mol) 

=> nH2 = nHCl/2 = 0.55 (mol) 

VH2 = 12.32 (l) 

BTKL : 

mA + mHCl = mX + mH2 

25.3 + 1.1*36.5 = mX + 1.1 

=> mX = 64.35 (g) 

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

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

a) Ta có: \(\left\{{}\begin{matrix}\Sigma n_{CaCl_2}=\dfrac{33,3}{111}=0,3\left(mol\right)\\n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{CaO}=0,1mol\\n_{CaCO_3}=0,2mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaO}=0,1\cdot56=5,6\left(g\right)\\m_{CaCO_3}=0,2\cdot100=20\left(g\right)\end{matrix}\right.\)

\(\Rightarrow m_{hh}=20+5,6=25,6\left(g\right)\)

b) Theo các PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(1\right)}=0,2mol\\n_{HCl\left(2\right)}=0,4mol\end{matrix}\right.\)

\(\Rightarrow\Sigma n_{HCl}=0,6mol\) \(\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,3}=2\left(M\right)\)

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

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

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

\(\Sigma n_{CaCl_2}=\dfrac{33,3}{111}=0,3\left(mol\right)\)

Theo PT (2): \(n_{CaCl_2}=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow n_{CaCl_2\left(1\right)}=0,3-0,2=0,1\left(mol\right)\)

Theo PT (1): \(n_{CaO}=n_{CaCl_2}=0,1\left(mol\right)\)

Theo PT (2): \(n_{CaCO_3}=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow m_A=m_{CaO}+m_{CaCO_3}=0,1.56+0,2.100=25,6\left(g\right)\)

b, Theo PT (1) + (2): \(\Sigma n_{HCl}=2n_{CaO}+2n_{CaCO_3}=0,6\left(mol\right)\)

\(\Rightarrow C_{M_{ddHCl}}=\dfrac{0,6}{0,3}=2M\)

Bạn tham khảo nhé!

nH2 = 2.24/22.4 = 0.1 (mol) 

Mg + 2HCl => MgCl2 + H2 

0.1___0.2___________0.1 

mMgO = 6.4 - 0.1*24 = 4 (g) 

nMgO = 4/40 = 0.1 (mol) 

MgO + 2HCl => MgCl2 + h2O 

0.1_____0.2 

nHCl = 0.2 + 0.2 = 0.4 (mol) 

V dd HCl = 0.4 / 0.8  = 0.5 (l) 

Gọi : \(\left\{{}\begin{matrix}n_{MgO}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\)⇒ 40a + 65b = 34(1)

\(MgO + 2HCl \to MgCl_2 + H_2O\\ Zn + 2HCl \to ZnCl_2 + H_2O\)

Muối gồm  :\(\left\{{}\begin{matrix}n_{MgCl_2}=a\left(mol\right)\\n_{ZnCl_2}=b\left(mol\right)\end{matrix}\right.\)

Suy ra : 95a + 136b = 73,4(2) 

Từ (1)(2) suy ra a = 0,2 ; b = 0,4

Vậy :

\(\%m_{MgO} = \dfrac{0,2.40}{34} .100\% = 23,53\%\\ \%m_{Zn} = 100\% - 23,53\% = 76,47\%\)

PTHH: \(2NaCl_{\left(rắn\right)}+H_2SO_{4\left(đ\right)}\underrightarrow{t^o}Na_2SO_4+2HCl\)

Ta có: \(n_{HCl}=\dfrac{50\cdot14,6\%}{36,5}=0,2\left(mol\right)=n_{NaCl}\)

\(\Rightarrow m_{NaCl}=0,2\cdot58,5=11,7\left(g\right)\)

Ta có: \(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\)

Gọi a và b lần lượt là số mol của Al và Fe

Bảo toàn mol e: \(3a+2b=1,4\)

                 Mà \(27a+56b=27,8\)

\(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,2\cdot27}{27,8}\cdot100\approx19,42\%\\\%m_{Fe}=80,58\%\end{matrix}\right.\)

Đặt : 

nAl = a mol 

nFe = b mol 

mB = 27a + 56b = 27.8 (g) (1) 

2Al + 6HCl => 2AlCl3 + 3H2 

a___________________1.5a

Fe + 2HCl => FeCl2 + H2 

b__________________b

nH2 = 1.5a + b = 15.68/22.4 = 0.7 (mol) (2) 

(1) , (2) : 

a = 0.2

b = 0.4 

%Al = 5.4/27.8 * 100% = 19.42%

%Fe = 100 - 19.42 = 80.58%