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

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

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

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

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

a)

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

Theo PTHH :

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

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

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ pthh:Mg+2HCl\rightarrow MgCl_2+H_2\) 
          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

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

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

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

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

b, Ta có: \(n_{H_2}=0,15\left(mol\right)\)

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

\(\Rightarrow m_{Al}=0,1.27=2,7\left(g\right)\)

\(\Rightarrow m_{Al_2O_3}=7,8-2,7=5,1\left(g\right)\)

c, Có: \(n_{Al_2O_3}=\dfrac{5,1}{102}=0,05\left(mol\right)\)

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

\(n_{AlCl_3}=n_{Al}+2n_{Al_2O_3}=0,2\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\Rightarrow m_{ddHCl}=\dfrac{21,9}{10\%}=219\left(g\right)\)

⇒ m _{dd sau pư} = 7,8 + 219 - 0,15.2 = 226,5 (g)

\(\Rightarrow C\%_{AlCl_3}=\dfrac{0,2.133,5}{226,5}.100\%\approx11,79\%\)

Bạn tham khảo nhé!

Em coi lại đề xem đúng chưa, chứ anh thấy cái thể tích khí số xấu lắm

nH2=4,968/22,4~0,22(mol)

A),PTHH:

Fe+2HCl→FeCl2+H2

B)nFe=nH2~0,22(mol)

⇒mFe=0,22.56~12,42(g)

⇒%Fe= 12,42 /20 .100 % = 62,1 %

⇒%Cu=100%−62,1%=37,9%

C)

nHCl=2nH2=0,44(mol)

⇒CnHCl=0,44/0,44=1M

$a)PTHH:2Al+6HCl\to 2AlCl_3+3H_2$

$n_{H_2}=\dfrac{5,04}{22,4}=0,225(mol)$

$\Rightarrow n_{Al}=0,15(mol)$

$\Rightarrow \%m_{Al}=\dfrac{0,15.27}{9,45}.100\%\approx 42,86\%$

$\Rightarrow \%m_{Cu}=100-42,86=57,14\%$

$b)$ Theo PT: $n_{HCl}=2n_{H_2}=0,45(mol)$

$\Rightarrow C_{M_{HCl}}=\dfrac{0,45.110\%}{0,5}=0,99M$

\(a)2Na+2HCl\xrightarrow[]{}2NaCl+H_2\\ 2K+2HCl\xrightarrow[]{}2KCl+H_2 \\ b)n_{H_2}=\dfrac{2,24}{22,4}=0,1mol\\ n_{Na}=a,n_K=b\\ \Rightarrow\left\{{}\begin{matrix}23a+39b=6,2\\\dfrac{1}{2}a+\dfrac{1}{2}b=0,1\end{matrix}\right.\\ \Rightarrow a=b=0,1mol\\ \%_{Na}=\dfrac{0,1.23}{6,2}\cdot100=37,1\%\\ \%_K=100-37,1=62,9\%\)