\(n_{H_2}=\dfrac{11,2}{22,4}=0,5mol\)

Gọi \(\left\{{}\begin{matrix}n_{Al}=x\\n_{Mg}=y\end{matrix}\right.\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

 x                                                 3/2x

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

 y                                           y ( mol )

Ta có:

\(\left\{{}\begin{matrix}27x+24y=10,2\\\dfrac{3}{2}x+y=0,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)

\(\Rightarrow m_{Al}=0,2.27=5,4g\)

\(\Rightarrow m_{Mg}=0,2.24=4,8g\)

=> Chọn C

\(n_{H_2}=\frac{5,6}{22,4}=0,25(mol)\\ m_{H_2}=0,25.2=0,5(g)\\ BT H:\\ n_{HCl}=2n_{H_2}=0,25.2=0,5(mol)\\ m_{HCl}=0,5.36,5=18,25(g)\\ BTKL:\\ m_{hh}+m_{HCl}=m_{muối}+m_{H_2}\\ 15+18,25=m_{muối}+0,5\\ \to m_{muối}=32,75(g)\\ \to D\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)

\(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\Rightarrow24x+65y=11,3\left(1\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

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

\(\Rightarrow x+y=0,3\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2mol\\y=0,1mol\end{matrix}\right.\)

a)\(\%m_{Mg}=\dfrac{0,2\cdot24}{11,3}\cdot100\%=42,48\%\)

\(\%m_{Zn}=100\%-42,48\%=57,52\%\)

b)\(n_{HCl}=2\left(n_{Mg}+n_{Zn}\right)=2\cdot\left(0,2+0,1\right)=0,6mol\)

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

Bài 1:

\(n_{HCl}=2.0,16=0,32\left(mol\right);n_{H_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)

PTHH: Mg + 2HCl → MgCl₂ + H₂

PTHH: Fe + 2HCl → FeCl₂ + H₂

\(m_{H_2}=0,16.2=0,32\left(g\right)\)

\(m_{HCl}=0,32.36,5=11,68\left(g\right)\)

Theo ĐLBTKL ta có: \(m_{MgCl_2+FeCl_2}=1,4+11,68-0,32=12,76\left(g\right)\)

Bài 12:

Theo ĐLBTKL, ta có:

\(m_{hhkl}+m_{O_2}=m_{hh.oxit}\\ \Leftrightarrow11,9+m_{O_2}=18,3\\ \Leftrightarrow m_{O_2}=18,3-11,9=6,4\left(g\right)\\ n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)\\ V_{O_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)

\(n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\)

\(\left\{{}\begin{matrix}Mg:x\left(mol\right)\\Al:y\left(mol\right)\end{matrix}\right.\)→ \(\left\{{}\begin{matrix}24x+27y=6,3\\x+1,5y=0,3\end{matrix}\right.\)→ \(\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)

Vậy :

\(\%m_{Mg} = \dfrac{0,15.24}{6,3}.100\% = 57,14\%\)

Câu 1: 

\(n_{H_2}=\dfrac{2.91362}{22.4}=0.13mol\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

 a           a                a           a

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

 2b          3b               b                 3b

Ta có: \(\left\{{}\begin{matrix}24a+54b=2.58\\a+3b=0.13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0.04\\b=0.03\end{matrix}\right.\)

\(m_{Mg}=0.04\times24=0.96g\)

\(m_{Al}=0.03\times2\times27=1.62g\)

\(V_{H_2SO_4}=\dfrac{0.04+3\times0.03}{0.5}=0.26l\)

Câu 2:

\(n_{H_2}=\dfrac{3.136}{22.4}=0.14mol\)

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

a            a              a             a

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b           b                 b         b

Ta có: \(\left\{{}\begin{matrix}24a+56b=4.96\\a+b=0.14\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=0.09\\b=0.05\end{matrix}\right.\)

\(m_{Mg}=0.09\times24=2.16g\)

\(m_{Fe}=0.05\times56=2.8g\)

\(C\%_{H_2SO_4}=\dfrac{0.14\times98\times100}{200}=6.86\%\)

Câu 3: 

\(n_{H_2}=\dfrac{1.568}{22.4}=0.07mol\)

\(Ba+H_2SO_4\rightarrow BaSO_4+H_2\)

a           a              a             a

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

b          b                 b             b

Ta có: \(\left\{{}\begin{matrix}137a+24b=3.94\\a+b=0.07\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=0.02\\b=0.05\end{matrix}\right.\)

\(m_{Ba}=0.02\times137=2.74g\)

\(m_{Mg}=0.05\times24=1.2g\)

\(CM_{H_2SO_4}=\dfrac{0.07}{0.1}=0.7M\)

a) 

Gọi số mol Mg, Al là a, b (mol)

=> 24a + 27b = 26,25 (1)

\(n_{H_2}=\dfrac{30,8}{22,4}=1,375\left(mol\right)\)

PTHH: Mg + 2HCl --> MgCl₂ + H₂

             a-->2a--------->a------>a

            2Al + 6HCl --> 2AlCl₃ + 3H₂

             b---->3b------->b------>1,5b

=> a + 1,5b = 1,375 (2)

(1)(2) => a = 0,25 (mol); b = 0,75 (mol)

=> \(\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,25.24}{26,25}.100\%=22,857\%\\\%m_{Al}=\dfrac{0,75.27}{26,25}.100\%=77,143\%\end{matrix}\right.\)

b)

n_HCl = 2a + 3b = 2,75 (mol)

=> m_HCl = 2,75.36,5 = 100,375 (g)

=> \(m_{dd.HCl}=\dfrac{100,375.100}{10}=1003,75\left(g\right)\)

c) 

m_{dd sau pư} = 1003,75 + 26,25 - 1,375.2 = 1027,25 (g)

\(\left\{{}\begin{matrix}C\%_{MgCl_2}=\dfrac{0,25.95}{1027,25}.100\%=2,312\%\\C\%_{AlCl_3}=\dfrac{0,75.133,5}{1027,25}.100\%=9,747\%\end{matrix}\right.\)