\(n_{Fe}=\dfrac{2.8}{56}=0.05\left(mol\right)\)

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

a) Chất tan : FeSO₄

Chất khí : H₂

\(m_{FeSO_4}=0.05\cdot152=7.6\left(g\right)\)

\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)

Fe + H2SO4 -> FeSO4 + H2

 1           1                 1            1

A là: FeSO4

B là: H2

nFe = 2,8/56 = 0,05 mol => nH2SO4 = 0,05 mol

mFeSO4 = 0,05 x (56+32+64) = 7,6 gam 

VH2 = 0,05 x 22,4 = 1,12 lit

\(n_{Al}=\dfrac{5.4}{27}=0.2\left(mol\right)\)

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

\(0.2..........0.3...............0.1...........0.3\)

\(m_{H_2SO_4}=0.3\cdot98=29.4\left(g\right)\)

\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)

\(m_{dd_{H_2SO_4}}=\dfrac{29.4\cdot100}{20}=147\left(g\right)\)

\(m_{Al_2\left(SO_4\right)_3}=0.1\cdot342=34.2\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng }}=5.4+147-0.3\cdot2=151.8\left(g\right)\)

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34.2}{151.8}\cdot100\%=22.53\%\)

Gọi $n_{H_2} = a(mol) \Rightarrow n_{HCl} = 2a(mol)$

Bảo toàn khối lượng : 

$m_A + m_{HCl} = m_{muối} + m_{H_2}$

Suy ra : $1,6 + 2a.36,5 = 2,31 + 2a$

Suy ra : $a = 0,01$

$V_{H_2} = 0,01.22,4 = 0,224(lít)$

a)

$Na_2O + H_2O \to 2NaOH + H_2$
$2Na + 2H_2O \to 2NaOH + H_2$
b)

$n_{Na} = 2n_{H_2} = 0,05.2 = 0,1(mol)$

$\Rightarrow n_{Na_2O} = \dfrac{14,7 - 0,1.23}{62} = 0,2(mol)$
$n_{NaOH} = n_{Na} + 2n_{Na_2O} = 0,5(mol)$
$C_{M_{NaOH}} = \dfrac{0,5}{0,2} = 2,5M$

c)

$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{H_2SO_4} = \dfrac{1}{2}n_{NaOH} = 0,25(mol)$

$m_{dd\ H_2SO_4} = \dfrac{0,25.98}{20\%} = 122,5(gam)$
$V_{dd\ H_2SO_4}  = \dfrac{122,5}{1,4} = 87,5(ml)$

Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\)

\(n_{CuO}=\dfrac{5}{80}=0,0625\left(mol\right)\)

PTHH:

Fe + H₂SO₄ ---> FeSO₄ + H₂

a ----------------------> a

Mg + H₂SO₄ ---> MgSO₄ + H₂

b ------------------------> b

MgSO₄ + 2KOH ---> Mg(OH)₂ + K₂SO₄

b ---------------------------> b

FeSO₄ + 2KOH ---> Fe(OH)₂ + K₂SO₄

a ---------------------------> a

4Fe(OH)₂ + O₂ + 2H₂O --t^o--> 4Fe(OH)₃

a ----------------------------------------> a

2Fe(OH)₃ --t^o--> Fe₂O₃ + 3H₂O

a --------------------> 0,5a

Mg(OH)₂ --t^o--> MgO + H₂O

b -------------------> b

2Cu + O₂ --t^o--> 2CuO

0,0625 <------------- 0,0625

Hệ pt \(\left\{{}\begin{matrix}56a+24b+0,0625.64=20\\160.0,5a+40b=24\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Mg}=0,2.24=4,8\left(g\right)\\m_{Cu}=0,625.64=4\left(g\right)\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{11,2}{20}=56\%\\\%m_{Mg}=\dfrac{4,8}{20}=24\%\\\%m_{Cu}=\dfrac{4}{20}=20\%\end{matrix}\right.\)

\(\left\{{}\begin{matrix}Fe:x\left(mol\right)\\Mg:y\left(mol\right)\end{matrix}\right.\)

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

x                           x             x

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

y                           y               y

Khí A nhận được là \(H_2\)

\(FeSO_4+2KOH\rightarrow Fe\left(OH\right)_2\downarrow+K_2SO_4\)

x                                     x

\(MgSO_4+2KOH\rightarrow Mg\left(OH\right)_2\downarrow+K_2SO_4\)

y                                       y                

\(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)

x                     \(\dfrac{x}{2}\)

\(Mg\left(OH\right)_2\underrightarrow{t^o}MgO+H_2O\)

y                     y

\(2Cu+O_2\underrightarrow{t^o}2CuO\)

\(n_{CuO}=\dfrac{5}{80}=0,0625mol\Rightarrow m_{Cu}=4g\)

\(m_{Fe+Mg}=20-4=16g\)

\(\Rightarrow\left\{{}\begin{matrix}56x+24y=16\\80x+40y=24\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)

\(\%m_{Fe}=\dfrac{0,2\cdot56}{20}\cdot100\%=56\%\)

\(\%m_{Mg}=\dfrac{0,2\cdot24}{20}\cdot100\%=24\%\)

\(\%m_{Cu}=100\%-\left(56\%+24\%\right)=20\%\)

\(a) Fe + 2HCl \to FeCl_2 + H_2\\ b) n_{H_2} = n_{Fe} = \dfrac{8,4}{56} = 0,15(mol)\\ V_{H_2} = 0,15.22,4 = 3,36(lít)\\ n_{HCl} = 2n_{H_2} = 0,3(mol)\ m_{HCl} = 0,3.36,5 = 10,95(gam)\)

\(n_{Fe}=\dfrac{8.4}{56}=0.15\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(0.15......0.3..................0.15\)

\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)

\(V_{H_2}=0.15\cdot22.4=3.36\left(l\right)\)

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

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

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

\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)

⇒ m_FeO = 12,6 - 5,4 = 7,2 (g)

c, Phần này đề cho dd NaOH dư hay vừa đủ bạn nhỉ?

d, Cho hh vào dd H₂SO₄ đặc nguội thì có khí thoát ra.

PT: \(2FeO+4H_2SO_{4\left(đ\right)}\rightarrow Fe_2\left(SO_4\right)_3+SO_2+4H_2O\)

Ta có: \(n_{FeO}=\dfrac{7,2}{72}=0,1\left(mol\right)\)

Theo PT: \(n_{SO_2}=\dfrac{1}{2}n_{FeO}=0,05\left(mol\right)\)

\(\Rightarrow V_{SO_2}=0,05.22,4=1,12\left(l\right)\)