\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)

        1          2            1           1

       0,2       0,4         0,2         0,2

      \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O|\)

          1            6               2              3

         0,2         1,2            0,4

\(n_{Fe}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{Fe}=0,2.56=11,2\left(g\right)\)

\(m_{Fe2O3}=27,2-11,2=16\left(g\right)\)

⁰/₀Fe = \(\dfrac{11,2.100}{27,2}=41,18\)⁰/₀

⁰/₀Fe₂O₃ = \(\dfrac{16.100}{27,2}=58,82\)⁰/₀

b) Có : \(m_{Fe2O3}=16\left(g\right)\)

 \(n_{Fe2O3}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,4+1,2=1,6\left(mol\right)\)

\(V_{HCl}=\dfrac{1,6}{2}=0,8\left(l\right)\)

c) \(n_{FeCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

   \(n_{FeCl3}=\dfrac{1,2.2}{6}=0,4\left(mol\right)\)

  \(C_{M_{FeCl2}}=\dfrac{0,2}{0,8}=0,25\left(M\right)\)

  \(C_{M_{FeCl3}}=\dfrac{0,4}{0,8}=0,5\left(M\right)\)

 Chúc bạn học tốt

PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

               a_______a________a______a                  (mol)

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

              b_______\(\dfrac{3}{2}\)b_________\(\dfrac{1}{2}\)b_____\(\dfrac{3}{2}\)b              (mol)

a) Ta lập HPT: \(\left\{{}\begin{matrix}24a+27b=8,25\\a+\dfrac{3}{2}b=\dfrac{2,24}{22,4}=0,1\end{matrix}\right.\)  \(\Leftrightarrow\) Hệ có nghiệm âm   

*Bạn xem lại đề !!!

thôi thì mình làm cho bn vậy, câu a ko làm dc đâu, làm câu b thôi, làm sao biết dc chất nào dư khi chỉ có số mol 1 chất?

nK2SO3=0.1367(mol)

mddH2SO4=Vdd.D=200.1,04=208(g)

K2SO3+H2SO4-->K2SO4+H2O+SO2

0.1367----0.1367----0.1367---------0.1367   (mol)

mddspu=100+208-0,1367.64=299.2512(g) ; mK2SO4=0,1367.174=23.7858(g)

==>C%=23.7858.100/299.512=7.94%

2)pt bn tự ghi nhé

ta có hệ pt: 56a+27b=11 và a+3b/2=8.96/22.4==>a=0.1, b=0.2

==>%Fe=0.1x56x100/11=50.9%

%Al=100%-50.9%=49.1%

b)nH2SO4= 0.7(mol)==>VddH2SO4=0.7/2=0.35(L)

bn xem lại đề nhé, bài 1mình thấy số mol ra lẻ><

a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{5,25}.100\%\approx51,43\%\\\%m_{Al_2O_3}\approx48,57\%\end{matrix}\right.\)

b, \(n_{Al_2O_3}=\dfrac{5,25-0,1.27}{102}=0,025\left(mol\right)\)

Theo PT: \(n_{HCl}=3n_{Al}+6n_{Al_2O_3}=0,45\left(mol\right)\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,45.36,5}{29,2\%}=56,25\left(g\right)\)

c, \(n_{H_2SO_4}=\dfrac{1}{2}n_{HCl}=0,225\left(mol\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,225.98}{19,6\%}=112,5\left(g\right)\)

Pthh: 

Mg+ H²SO⁴(loãng)-> MgSO⁴ + H²

Fe + H²SO⁴(loãng)-> FeSO⁴ +H²

Mọi người giúp mình vs

\(n_{HCl} = \dfrac{448.1,12.3,65\%}{36,5} = 0,50176(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ ZnO + 2HCl \to ZnCl_2 + H_2O\\ n_{Zn} = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)\\ n_{ZnO} = \dfrac{n_{HCl} - 2n_{Zn}}{2} = \dfrac{0,50176-0,1.2}{2} = 0,15088(mol)\\ \%m_{Zn} = \dfrac{0,1.65}{0,1.65 + 0,15088.81}.100\% = 34,72\%\\ \%m_{ZnO} = 65,28\%\)

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

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)

\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)

c, \(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Mg}+n_{MgO}=0,15\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{14,7}{19,6\%}=75\left(g\right)\)