nH2=0,1(mol)

PTHH: Mg + 2 HCl -> MgCl2 + H2

0,1__________0,2___________0,1(mol)

MgO + 2 HCl -> MgCl2 + H2O

0,05____0,1___0,05(mol)

mMg=0,1. 24= 2,4(g) -> mMgO=4,4-2,4= 2(g) -> nMgO=0,05((mol)

b) %mMg= (2,4/4,4).100=54,545%

=> %mMgO=45,455%

c) nHCl=0,3(mol) -> mHCl=0,3.36,5=10,95(g)

=> mddHCl=(10,95.100)/7,3=150(g)

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
          0,2                                     0,2 
 \(MgO+2HCl\rightarrow MgCl_2+H_2O\)
\(m_{Mg}=0,2.24=4,8g\\ m_{MgO}=18,4-4,8=13,6g\)

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

\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)

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

Theo PT: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}=\dfrac{0,1.24}{34,4}.100\%\approx6,98\%\\\%m_{Fe_2O_3}\approx93,02\%\end{matrix}\right.\)

c, Ta có: \(m_{Fe_2O_3}=34,4-0,1.24=32\left(g\right)\Rightarrow n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{MgSO_4}=n_{Mg}=0,1\left(mol\right)\\n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,2\left(mol\right)\end{matrix}\right.\)\(n_{H_2SO_4}=n_{Mg}+3n_{Fe_2O_3}=0,7\left(mol\right)\Rightarrow m_{H_2SO_4}=0,7.98=68,6\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{68,6}{10\%}=686\left(g\right)\)

Ta có: m _{dd sau pư} = 34,4 + 686 - 0,1.2 = 720,2 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgSO_4}=\dfrac{0,1.120}{720,2}.100\%\approx1,67\%\\C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,2.400}{720,2}.100\%\approx11,11\%\end{matrix}\right.\)

Bạn tham khảo nhé!

Giúp mình với 

a) H2SO4 + Mg → MgSO4 +  H2  (1)

Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O (2)

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

=> \(\%m_{Mg}=\dfrac{0,1.24}{34,4}.100=6,98\%\)

\(\%m_{Fe_2O_3}=\dfrac{32}{34,4}.100=93,02\%\)

c) \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

\(m_{ddsaupu}=34,4+\dfrac{\left(0,1+0,2.3\right).98}{10\%}-0,1.2=720,2\left(g\right)\)

Dung dịch B : MgSO4,  Fe2(SO4)3

\(C\%_{MgSO_4}=\dfrac{0,1.120}{720,2}.100=1,67\%\)

\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,2.400}{720,2}.100=11,11\%\)

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

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

\(n_{Mg}=n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

\(m_{Mg}=0.05\cdot24=1.2g\)

\(m_{MgO}=9.2-1.2=8\left(g\right)\)

a)

$Mg + 2HCl \to MgCl_2 + H_2$
$MgO + 2HCl \to MgCl_2 + H_2O$
b)

Theo PTHH :

$n_{Mg} = n_{H_2} = \dfrac{1,12}{22,4} = 0,05(mol)$

$m_{Mg} = 0,05.24 = 1,2(gam)$
$m_{MgO} = 9,2 - 1,2 = 8(gam)$

a) Fe + 2HCl --> FeCl₂ + H₂

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

_____0,15<-0,3<----0,15<---0,15

\(\left\{{}\begin{matrix}\%Fe=\dfrac{0,15.56}{12}.100\%=70\%\%\\\%Cu=100\%-70\%=30\%\end{matrix}\right.\)

c) m_HCl = 0,3.36,5 = 10,95 (g)

=> \(m_{dd}=\dfrac{10,95.100}{10}=109,5\left(g\right)\)

d) mdd  = 12 + 109,5 - 0,15.2 = 121,2 (g)

 \(C\%\left(FeCl_2\right)=\dfrac{0,15.127}{121,2}.100\%=15,718\%\)

a: \(Cu+2HCl->CuCl_2+H_2\)

\(Fe+2HCl->FeCl_2+H_2\)

a) Fe + 2HCl --> FeCl₂ + H₂

FeO + 2HCl --> FeCl₂ + H₂O

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

PTHH: Fe + 2HCl --> FeCl₂ + H₂

______0,5<-1<------0,5<---0,5

=> m_Fe = 0,5.56 = 28 (g)

=> \(\left\{{}\begin{matrix}\%Fe=\dfrac{28}{100}.100\%=28\%\\\%FeO=100\%-28\%=72\%\end{matrix}\right.\)

c) \(n_{FeO}=\dfrac{72}{72}=1\left(mol\right)\)

PTHH: FeO + 2HCl --> FeCl₂ + H₂O

______1---->2

=> m_HCl = (1+2).36,5 = 109,5 (g)

=> \(m_{ddHCl}=\dfrac{109,5.100}{30}=365\left(g\right)\)

=> \(V_{ddHCl}=\dfrac{365}{1,15}=317,39\left(ml\right)\)

Zn+H2SO4->ZnSO4+H2

2Al+3H2SO4->Al2(SO4)3+3H2

Fe+H2SO4->FeSO4+H2

m muối=26,7+\(\dfrac{16,8}{22,4}\).96=98,7g

a)

$Fe + 2HCl \to FeCl_2 + H_2$
b)

Theo PTHH : $n_{Fe} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$
$\Rightarrow m_{Fe} = 0,2.56 = 11,2(gam)$
$\Rightarrow m_{Cu} = 15,6 - 11,2  = 4,4(gam)$