a. \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

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

0,25 ..... 0,5 ................... 0,25 (mol) 

\(m_{Mg}=0,25.24=6\left(g\right)\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{6}{10}.100\%=60\%\\\%m_{MgO}=100\%-60\%=40\%\end{matrix}\right.\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

0,1 ....... 0,2 (mol) 

\(n_{HCl}=0,25+0,1=0,35\left(mol\right)\)

\(C_M\left(HCl\right)=\dfrac{0,35}{0,1}=3,5\left(M\right)\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ Mg+2HCl\to MgCl_2+H_2\\ MgO+2HCl\to MgCl_2+H_2O\\ \Rightarrow n_{Mg}=0,1(mol)\\ \Rightarrow \%_{Mg}=\dfrac{0,1.24}{6,4}.100\%=37,5\%\\ \Rightarrow \%_{MgO}=100\%-37,5\%=62,5\%\)

\(b,n_{MgO}=\dfrac{6,4-0,1.24}{40}=0,1(mol)\\ \Rightarrow n_{HCl}=2.0,1+2.0,1=0,4(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,4}{0,5}=0,8(l)\\ c,n_{MgCl_2}=0,1+0,1=0,2(mol)\\ \Rightarrow C_{M_{MgCl_2}}=\dfrac{0,2}{0,8}=0,25M\)

Chọn A

1.1. Al + NaOH + H2O ==> NaAlO2 + 3/2H2

nH2(1)=3,36/22,4=0.15(mol)

=> nAl(1)= nH2(1):3/2= 0.15:3/2= 0.1(mol)

2.Mg + 2HCl ==> MgCl2 + H2

3.2Al + 6HCl ==> 2AlCl3 + 3H2

4.Fe + 2HCl ==> FeCl2 + H2

=> \(n_{H_2\left(2,3,4\right)}=\) 10.08/22.4= 0.45(mol)

=> nH2(3)=0.1*3/2=0.15(mol)

MgCl2 + 2NaOH ==> Mg(OH)2 + 2NaCl

AlCl3 + 3NaOH ==> Al(OH)3 + 3NaCl

FeCl2 + 2NaOH ==> Fe(OH)2 + 2NaCl

\(n_{CO_3^{2-}}=0,1\left(mol\right);n_{HCO_3^-}=0,2\left(mol\right);n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(CO_3^{2-}+H^+\text{→}HCO_3^-\)

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

\(HCO_3^-+H^+\text{→}CO_2+H_2O\)

0,15        0,15      0,15               (mol)

Suy ra : 

\(V=\dfrac{0,1+0,15}{1}=0,25\left(lít\right)\)

Bài1

Fe +2HCl----> FeCl2 +H2

Ta có

m\(_{HCl}=\frac{14,66.200}{100}=29,32\left(g\right)\)

n\(_{HCl}=\frac{29,32}{36,5}=0,8\left(mol\right)\)

Theo pthh

n\(_{Fe}=\frac{1}{2}n_{HCl}=0,4\left(mol\right)\)

m=m\(_{Fe}=0,4.56=22,4\left(g\right)\)

Theo pthh

n\(_{FeCl2}=n_{Fe}=0,4\left(mol\right)\)

m\(_{FeCl2}=0,4.127=50,8\left(g\right)\)

m\(_{H2}=0,8\left(g\right)\)

mdd= 22,4+200=0,8=221,6(g)

C%=\(\frac{50,8}{221,6}.100\%=22,92\%\)

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Bài 2

2K+2H2O--->2KOH+H2

Ta có

n\(_K=\frac{15,6}{137}=0,1\left(mol\right)\)

Theo pthh

n\(_{H2}=\frac{1}{2}n_K=0,05\left(mol\right)\)

m\(_{H2}=0,1\left(g\right)\)

m=m\(_{H2O}=200+0,1-15,6=184,5\left(g\right)\)

V\(_{H2}=0,05.22,4=1,12\left(l\right)\)

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