Em đăng sang môn Hoá nha

\(2Al+6HCl->2AlCl_3+3H_2\\ Fe+2HCl->FeCl_2+H_2\\ n_{Al}=a;n_{Fe}=b\\ 27a+56b=8,3\\ 1,5a+b=\dfrac{5,6}{22,4}=0,25\\ a=b=0,1\\ m_{Al}=27\cdot0,1=2,7g\\ m_{Fe}=8,3-2,7=5,6g\\ a=\dfrac{3a+2b}{500}\cdot36,5=3,65\%\\ m_{ddsau}=508,3-0,25\cdot2=507,8g\\ C\%_{AlCl_3}=\dfrac{133,5a}{507,8}=2,63\%\\ C\%_{FeCl_2}=\dfrac{127b}{507,8}=2,50\%\)

Đặt: \(\left\{{}\begin{matrix}x=n_{Fe}\left(mol\right)\\y=n_{Al}\left(mol\right)\end{matrix}\right.\)

\(\sum m_{hh}=11\left(g\right)\Rightarrow56x+27y=11\left(1\right)\)

\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\\ \left(mol\right)....x\rightarrow..2x........x......x\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\\ \left(mol\right)....y\rightarrow..3y.........y......1,5y\)

\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ \Rightarrow x+1,5y=0,4\left(2\right)\)

\(\xrightarrow[\left(2\right)]{\left(1\right)}\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

a) \(\%m_{Fe}=\dfrac{56.0,1}{11}=51\%\)

\(\rightarrow\%m_{Al}=49\%\)

b) \(\sum m_{ctHCl}=\left(2.0,1+3.0,2\right).36,5=29,2\left(g\right)\)

\(m_{ddHCl}=\dfrac{29,2.100\%}{10\%}=292\left(g\right)\)

c) \(m_{H_2\uparrow}=\left(1.0,1+1,5.0,2\right).2=0,8\left(g\right)\)

\(m_{ddsaupu}=m_{hh}+m_{ddHCl}-m_{H_2\uparrow}=11+292-0,8=302,2\left(g\right)\)

\(C\%_{FeCl_2}=\dfrac{0,1.127}{302,2}.100=4,2\%\\ C\%_{AlCl_3}=\dfrac{0,2.133,5}{302,2}.100=8,8\%\)

Đặt: {x=nFe(mol)y=nAl(mol){x=nFe(mol)y=nAl(mol)

∑mhh=11(g)⇒56x+27y=11(1)∑mhh=11(g)⇒56x+27y=11(1)

%mFe=56.0,111=51%%mFe=56.0,111=51%

→%mAl=49%→%mAl=49%

b) ∑mctHCl=(2.0,1+3.0,2).36,5=29,2(g)∑mctHCl=(2.0,1+3.0,2).36,5=29,2(g)

Đặt nMg = x, nCa = y

=> 24x + 40y = 6,4 (1)

Pt: 

Mg + H₂SO₄ => MgSO₄ + H₂

x                   =>       x          x

Ca + H₂SO₄ => CaSO₄ + H₂

y                    =>      y           y

=> x + y = \(\dfrac{4,48}{22,4}=0,2\) (2)

Từ 1 và 2 => x = 0,x và y = 0,1

=> mMg = 0,1.24 = 2,4

mCa = 6,4 - 2,4 = 4g

m muối = mMgSO₄ + mCaSO₄ = 0,1.120 + 0,1.136 = 25,6g

a, PTHH:

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

\(Ca+H_2SO_4\rightarrow CaSO_4+H_2\)

Bảo toàn e: \(n_{Mg}+n_{Ca}=n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Lại có: \(24n_{Mg}+40n_{Ca}=6,4\)

\(\Rightarrow n_{Mg}=n_{Ca}=0,1\left(mol\right)\)

\(\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right);n_{Ca}=0,1.40=4\left(g\right)\)

b, Tính kim loại muối??

a, \(m_{hh}=m_{Al}+m_{Fe}=27n_{Al}+56n_{Fe}=11,1\left(I\right)\)

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

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PTHH : \(\dfrac{3}{2}n_{Al}+n_{Fe}=n_{H2}=0,3\left(II\right)\)

- Giair 1 và 2 => \(\left\{{}\begin{matrix}n_{Al}=0,1\\n_{Fe}=0,15\end{matrix}\right.\) mol

\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=2,7g\left(24,32\%\right)\\m_{Fe}=8,4g\left(75,68\%\right)\end{matrix}\right.\)

b, - Theo PTHH : \(n_{H2SO4du}=n_{H2SO4}-n_{H2SO4pu}=0,325mol\)

\(\Rightarrow m_{H2SO4du}=31,85g\)

Ta có ; \(m_{dd}=m_{ddH2SO4}+m_{hh}-m_{H2}=255,5g\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{H2SO4}=\dfrac{m}{mdd}.100\%=12,46\%\\C\%_{Al2\left(SO4\right)3}=\dfrac{m}{mdd}.100\%=6,7\%\\C\%_{FeSO4}=\dfrac{m}{mdd}.100\%=8,9\%\end{matrix}\right.\)

Vậy ...

a,nH2=\(\dfrac{6,72}{22,4}\)=0,3(mol)

Gọi số mol của Al là x, số mol của Fe là y

Ta có :

27x+56y=11,1 (1) 

1,5a+b=0,3 (2)

Từ (1),(2) ⇒x=0,1 ; y=0,15

%mAl=\(\dfrac{0,1.27}{11,1}.100\)=24,32%

%mFe=100%−24,32%=75,68%

b,nH2SO4=\(\dfrac{245.25\%}{98}\)=0,625(mol)

⇒nH2SO4.trong.Y=0,625−0,3=0,325(mol)

mdd(spu)=11,1+245−0,3.2=255,5(g)

nAl2(SO4)3=0,05(mol)

nFeSO4=0,15(mol)

⇒C%H2SO4=12,47%

C%Al2(SO4)3=6,2%

C%FeSO4=8,92%

a/ Gọi số mol Al, Mg trong hỗn hợp là a, b

PTHH:

2Al + 6HCl ===> 2AlCl₃ + 3H₂

a............................................1,5a

Mg + 2HCl ===> MgCl₂ + H₂

b.........................................b

n_H2 = 5,6 / 22,4 = 0,25 (mol)

The đề ra, ta có hệ phương trình:

\(\begin{cases}27a+24b=5,1\\1,5a+b=0,25\end{cases}\)=> \(\begin{cases}a=0,1\\b=0,1\end{cases}\)

=> m_Al = 0,1 x 27 = 2,7 gam

m_Mg = 0,1 x 24 = 2,4 gam

=> %m_Al = \(\frac{2,7}{5,1}.100\%=52,94\%\)

%m_Mg = 100% - 52,94% = 46,06%

b/ Tổng số mol của HCl = 0,3 + 0,2 = 0,5 mol

=> m_HCl = 0,5 x 36,5 = 18,25 gam

c/ Áp dụng định luật bảo toàn khối lượng, ta có
m_{hỗn hợp muối} = m_{kim loại} + m_HCl - m_H2

₌ 5,1 + 18,25 - 0,25 x 2 = 22,85 gam

\(\text{Đặt }\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\\ n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ a,PTHH:\left\{{}\begin{matrix}2Al+6HCl\rightarrow2AlCl_3+3H_2\\Fe+2HCl\rightarrow FeCl_2+H_2\end{matrix}\right.\\ b,\text{Theo đề ta có HPT: }\left\{{}\begin{matrix}27x+56y=8,3\\\dfrac{3}{2}x+y=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%_{Al}=\dfrac{0,1\cdot27}{8,3}\approx32,53\%\\\%_{Fe}\approx67,47\%\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}n_{AlCl_3}=0,1\left(mol\right)\\n_{FeCl_2}=0,1\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{AlCl_3}=0,1\cdot133,5=13,35\left(g\right)\\m_{FeCl_2}=0,1\cdot127=12,7\left(g\right)\end{matrix}\right.\\ \Rightarrow\sum m_{muối}=13,35+12,7=26,05\left(g\right)\)