PTHH: \(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

Ta có: \(n_{CuO}=\dfrac{29,4}{80}=0,3675\left(mol\right)=n_{CuSO_4}=n_{H_2SO_4}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{CuSO_4}=0,3675\cdot160=58,8\left(g\right)\\m_{H_2SO_4}=0,3675\cdot98=36,015\left(g\right)\\V_{H_2SO_4}=\dfrac{0,3675}{1}=0,3675\left(l\right)=367,5\left(ml\right)\end{matrix}\right.\)

a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, Ta có: \(n_{Fe}=\dfrac{19,6}{56}=0,35\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Fe}=0,35\left(mol\right)\Rightarrow V_{H_2}=0,35.22,4=7,84\left(l\right)\)

c, \(n_{H_2SO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{FeSO_4}=n_{Fe}=0,35\left(mol\right)\Rightarrow m_{FeSO_4}=0,35.152=53,2\left(g\right)\)

e, \(C_{M_{FeSO_4}}=\dfrac{0,35}{0,2}=1,75\left(M\right)\)

d, \(n_{H_2SO_4}=0,25.1,6=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{n_{Fe}}{1}< \dfrac{n_{H_2SO_4}}{1}\), ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{Fe}=0,35\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,4-0,35=0,05\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4\left(dư\right)}=0,05.98=4,9\left(g\right)\)

Cu(OH)₂ + H₂SO₄ \(\rightarrow\) CuSO₄ + 2H₂O

nCu(OH)₂ = \(\dfrac{29,4}{98}=0,3mol\)

Theo pt: nH₂SO₄ = nCu(OH)₂ = 0,3 mol

=> mH₂SO₄ = 0,3.98 = 29,4g

VH₂SO₄ = 0,3:1 = 0,3l

a)\(n_{Al}=\dfrac{18}{27}=\dfrac{2}{3}mol\)

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

\(\dfrac{2}{3}\)                                                 1

\(V_{H_2}=1\cdot22,4=22,4l\)                 

b)\(n_{Fe_2O_3}=\dfrac{32}{160}=0,2mol\)   

\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)

0,2            1          0,4      0,6

\(m_{Fe}=0,4\cdot56=22,4g\)

c)\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

    0,4     \(\dfrac{4}{15}\)

\(V_{O_2}=\dfrac{4}{16}\cdot22,4=\dfrac{448}{75}l\)

\(V_{kk}=5V_{O_2}=\dfrac{448}{15}l\approx29,87l\)

a)

$n_{Al} = 0,3(mol)$

$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$

Theo PTHH : 

$n_{H_2SO_4} = \dfrac{3}{2}n_{Al} = 0,45(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,45.98}{12,25\%} = 360(gam)$

b)

$n_{H_2} = n_{H_2SO_4} = 0,45(mol)$
$V_{H_2} = 0,45.22,4 = 10,08(lít)$

c)

$n_{Al_2(SO_4)_3} = 0,15(mol)$
$m_{dd\ sau\ pư} = 8,1 + 360 - 0,45.2 = 367,2(gam)$
$C\%_{Al_2(SO_4)_3} = \dfrac{0,15.342}{367,2}.100\% = 14\%$

\(\left(a\right)2Al+3H_2O\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ \left(b\right)n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\\ n_{Al}=\dfrac{0,6.2}{3}=0,4mol\\ m_{Al}=0,4.27=10,8g\\ \left(c\right)n_{O_2}=\dfrac{4,8}{32}=0,15mol\\ 4Al+3O_2\underrightarrow{t^0}2Al_2O_3\\ \Rightarrow\dfrac{0,4}{4}>\dfrac{0,15}{3}\Rightarrow Al.dư\\ n_{Al_2O_3}=\dfrac{0,15.2}{3}=0,1mol\\ m_{oxit}=m_{Al_2O_3}=0,1.102=10,2g\)

a: \(2Al+3H_2SO_4\rightarrow1Al_2\left(SO_4\right)_3+3H_2\uparrow\)

    0,4           0,6                0,2             0,6

b: \(n_{H_2}=\dfrac{13.44}{22.4}=0.6\left(mol\right)\)

=>\(n_{Al}=0.4\left(mol\right)\)

\(m_{Al}=0.4\cdot27=10.8\left(g\right)\)

c: \(4Al+3O_2\rightarrow2Al_2O_3\)

0,4                      0,2

\(m_{Al_2O_3}=0.2\left(27\cdot2+16\cdot3\right)=0.2\cdot102=20.4\left(g\right)\)

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

PTHH ;

2Al + 3H2SO4 --> Al2(SO4)3 + 3H2↑

  0,4           0,6               0,2           0,6

4Al + 3O2 ---> 2Al2O3

0,2       0,15          0,1 

\(n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\)

\(\dfrac{0,4}{4}>\dfrac{0,15}{3}\)

--> Tính theo oxi

\(b,m_{Al}=0,4.27=10,8\left(g\right)\)

\(c,m_{Al_2O_3}=0,1.102=10,2\left(g\right)\)

\(a,PTHH:2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ \Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\\ \Rightarrow m_{Al}=0,1\cdot27=2,7\left(g\right)\\ b,n_{H_2SO_4}=n_{H_2}=0,15\left(mol\right)\\ \Rightarrow m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\\ c,n_{Al_2\left(SO_4\right)_3}=\dfrac{1}{3}n_{H_2}=0,05\left(mol\right)\\ \Rightarrow m_{Al_2\left(SO_4\right)_3}=0,05\cdot342=17,1\left(g\right)\)

a) \(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)

PTHH: Al₂O₃ + 6HCl ---> 2AlCl₃ + 3H₂O

           0,2----->1,2-------->0,4

b) \(\left\{{}\begin{matrix}m_{\text{ax}it}=m_{HCl}=1,2.36,5=43,8\left(g\right)\\m_{mu\text{ố}i}=m_{AlCl_3}=0,4.133,5=53,4\left(g\right)\end{matrix}\right.\)