Câu 5 : 

\(n_{H2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)

            1           2             1           1

           0,1       0,2           0,1         0,1

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

             1           2              1             1

           0,15       0,3          0,15

a) \(n_{Mg}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(m_{Mg}=0,1.24=2,4\left(g\right)\)

\(m_{MgO}=8,4-2,4=6\left(g\right)\)

⁰/₀Mg = \(\dfrac{2,4.100}{8,4}=28,57\)⁰/₀

⁰/₀MgO = \(\dfrac{6.100}{8,4}=71,43\)⁰/₀

b) Có : \(m_{MgO}=6\left(g\right)\)

\(n_{MgO}=\dfrac{6}{40}=0,15\left(mol\right)\)

\(n_{HCl\left(tổng\right)}=0,2+0,3=0,5\left(mol\right)\)

\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)

\(m_{ddHCl}=\dfrac{18,25.100}{3,65}=500\left(g\right)\)

\(n_{MgCl2\left(tổng\right)}=0,1+0,15=0,25\left(mol\right)\)

⇒ \(m_{MgCl2}=0,15.95=14,25\left(g\right)\)

\(m_{ddspu}=8,4+500-\left(0,1.2\right)=508,2\left(g\right)\)

\(C_{MgCl2}=\dfrac{14,25.100}{508,2}=2,8\)⁰/₀

 Chúc bạn học tốt

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

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

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

\(0.05..............................0.05\)

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

\(m_{MgO}=9.5-1.2=8.3\left(g\right)\)

\(\%Mg=\dfrac{1.2}{9.5}\cdot100\%=12.63\%\)

\(\%MgO=100-12.63=87.36\%\)

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

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

a, Giả sử: \(\left\{{}\begin{matrix}n_{Mg}=x\left(mol\right)\\n_{Al}=y\left(mol\right)\end{matrix}\right.\)

⇒ 24x + 27y = 12,6 (1)

Ta có: \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Mg}+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}y\left(mol\right)\)

\(\Rightarrow x+\dfrac{3}{2}y=0,6\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,3\left(mol\right)\\y=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{MG}=\dfrac{0,3.24}{12,6}.100\%\approx57,1\%\\\%m_{Al}\approx42,9\%\end{matrix}\right.\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{H_2SO_4}=n_{H_2}=0,6\left(mol\right)\\n_{MgSO_4}=n_{Mg}=0,3\left(mol\right)\\n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{H_2SO_4}=0,6.98=58,8\left(g\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{58,8}{14,7\%}=400\left(g\right)\)

Ta có: m _{dd sau pư} = 12,6 + 400 - 0,6.2 = 411,4 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{MgSO_4}=\dfrac{0,3.120}{411,4}.100\%\approx8,75\%\\C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,1.342}{411,4}.100\%\approx8,31\%\end{matrix}\right.\)

Bạn tham khảo nhé!

1. Gọi mol của Mg và Al là x, y mol

=> 24x + 27y = 12,6 (1)

nH2 = 0,6 mol => x + 1,5y = 0,6 (2)

Từ (1) (2) => x = 0,3 ; y = 0,2 

=> %Mg = 57,14%

=> %Al = 42,86%

nH2=13,44/22,4=0,6(mol)

Đặt: nMg=a(mol); nAl=b(mol) (a,b>0)

1) PTHH: Mg + H2SO4 -> MgSO4 + H2

a__________a________a_____a(mol)

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

b___1,5b______0,5b____1,5b(mol)

Ta có hpt:

\(\left\{{}\begin{matrix}24a+27b=12,6\\a+1,5b=0,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\end{matrix}\right.\)

=> mMg=0,3.24=7,2(g)

=>%mMg= (7,2/12,6).100=57,143%

=>%mAl=42,857%

2) mMgSO4=120.a=120.0,3=36(g)

mAl2(SO4)3=342.0,5b=342.0,5.0,2= 34,2(g)

mH2SO4= (0,3+0,2.1,5).98=58,8(g)

=>mddH2SO4=58,8: 14,7%=400(g)

=>mddsau= 12,6+400 - 2.0,6= 411,4(g)

=>C%ddAl2(SO4)3= (34,2/411,4).100=8,313%

C%ddMgSO4=(36/411,4).100=8,751%

Ta có: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH:

Mg + 2HCl ---> MgCl₂ + H₂ (1)

MgO + 2HCl ---> MgCl₂ + H₂O (2)

Theo PT₍₁₎: \(n_{Mg}=n_{H_2}=0,05\left(mol\right)\)

=> \(m_{Mg}=0,05.24=1,2\left(g\right)\)

=> \(\%_{m_{Mg}}=\dfrac{1,2}{9,2}.100\%=13,04\%\)

\(\%_{m_{MgO}}=100\%-13,04\%=86,96\%\)

a, PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

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

Ta có: \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,4.27}{14,8}.100\%\approx72,97\%\\\%m_{MgO}\approx27,03\%\end{matrix}\right.\)

b, Ta có: \(n_{MgO}=\dfrac{14,8-0,4.27}{40}=0,1\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{H_2}+n_{MgO}=0,7\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,7.98=68,6\left(g\right)\)

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

c, Theo PT: \(\left\{{}\begin{matrix}n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{2}n_{Al}=0,2\left(mol\right)\\n_{MgSO_4}=n_{MgO}=0,1\left(mol\right)\end{matrix}\right.\)

Ta có: m _{dd sau pư} = 14,8 + 686 - 0,6.2 = 699,6 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{Al_2\left(SO_4\right)_3}=\dfrac{0,2.342}{699,6}.100\%\approx9,78\%\\C\%_{MgSO_4}=\dfrac{0,1.120}{699,6}.100\%\approx1,72\%\end{matrix}\right.\)