Ta có n_Al = \(\dfrac{0,54}{27}\) = 0,02 ( mol )

n_H2SO4 = \(\dfrac{9,8}{98}\) = 0,1 ( mol )

2Al + 3H₂SO₄ \(\rightarrow\) Al₂(SO₄)₃ + 3H₂

0,02....0,1

=> Lập tỉ số \(\dfrac{0,02}{2}:\dfrac{0,1}{3}\) = 0,01 < \(\dfrac{1}{30}\)

=> Sau phản ứng Al hết ; H₂SO₄ dư

=> V_H2 = 0,03 . 22,4 = 0,672 ( lít )

=> m_{H2SO4 dư} = ( 0,1 - 0,03 ) .98 = 6,86 ( gam )

\(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{0,98}{98}=0,01\left(mol\right)\)

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

0,02 0,01 ---------------------------> 0,01

Lập tỉ số: \(n_{Al}:n_{H_2SO_4}=\dfrac{1}{100}>\dfrac{1}{300}\)

=> Al dư, H₂SO₄ hết

\(V_{H_2}=0,01.22,4=0,224\left(l\right)\)

\(m_{Al\left(dư\right)}=\left(0,02-\dfrac{1}{150}\right).27=0,36\left(g\right)\)

a, Ta có n_Al = \(\dfrac{0,54}{27}\) = 0,02 ( mol )

n_H2SO4 = \(\dfrac{0,98}{98}\) = 0,01 ( mol )

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

0,02 0,01

=> Lập tỉ số \(\dfrac{0,02}{2}\) : \(\dfrac{0,01}{3}\) = 0,01 > \(\dfrac{1}{300}\)

=> Sau phản ứng Al còn dư

H₂SO₄ hết

2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

\(\dfrac{1}{150}\) \(\leftarrow\)0,01-------------------> 0,01

=> V_H2 = 0,01 . 22,4 = 0,224 ( lít )

b, Al còn dư

=> m_{Al dư} = ( 0,02 -\(\dfrac{1}{150}\) ) . 27 = 0,36 ( gam )

\(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\)

2Al + 3H₂SO₄ \(\rightarrow\) Al₂(SO₄)₃ + 3H₂

de: 0,02 \(\rightarrow\) 0,03

a, \(V_{H_2}=22,4.0,03=0,672l\)

\(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

Ta co: \(\dfrac{0,2}{1}>\dfrac{0,03}{1}\Rightarrow\) CuO dư

b, CuO + H₂ \(\rightarrow\) Cu + H₂O

de: 0,2 0,03

pu: 0,03 0,03 0,03 0,03

spu: 0,17 0 0,03 0,03

\(m_{Cu}=0,03.64=1,92g\)

$n_{Zn} = \dfrac{6,5}{65} = 0,1(mol) \\ PTHH: Zn + 2HCl \to ZnCl_2 + H_2 \\$$n_{H_2} = n_{Zn} = 0,1(Mol) \\ V_{H_2} = 0,1.22,4 = 2,24l \\b) PTHH: H_2 + CuO \xrightarrow[]{t^o} Cu + H_2O \\ n_{CuO} = \dfrac{12}{64} = 0,15(mol) \\ \to CuO dư$ $\\ n_{H_2} = n_{Cu} = 0,1(mol \\ m_{Cu} = 0,1.64 = 6,4(gam)$

Zn+2HCl->ZnCl2+H2

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

n Zn=0,1 mol

=>VH2=0,1.22,4=2,24l

H2+CuO-to>Cu+H2O

          0,15-----0,15

n CuO=0,15 mol

=>H2 dư

=>m Cu=0,15.64=9,6g

a,\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right);n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂

Mol:    0,1         0,15            0,05             0,15

b,Ta có: \(\dfrac{0,1}{2}< \dfrac{0,3}{3}\) ⇒ Al hết, H₂SO₄ dư

\(\Rightarrow m_{H_2SO_4dư}=\left(0,3-0,15\right).98=14,7\left(g\right)\)

c, \(m_{Al_2\left(SO_4\right)_3}=0,05.342=17,1\left(g\right)\)

d, \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)

nZn = 19.5/65 = 0.3 (mol) 

Zn + H2SO4 => ZnSO4 + H2 

0.3........................0.3.........0.3

VH2 = 0.3*22.4 = 6.72 (l) 

mZnSO4 = 0.3*161 = 48.3 (g) 

nCuO = 16/80 = 0.2 (mol) 

CuO + H2 -to-> Cu + H2O 

0.2........0.2 

=> H2 dư 

mH2 (dư) = ( 0.3 - 0.2 ) * 2 = 0.2 (g) 

nZn=0,3(mol)

a) PTHH: Zn + H2SO4 -> ZnSO4+ H2

0,3___________________0,3____0,3(mol)

mZnSO4=161.0,3=48,3(g)

b) V(H2,đktc)=0,3.22,4=6,72(l)

c) nCuO=16/80=0,2(mol)

PTHH: CuO + H2 -to-> Cu + H2O

vì: 0,3/1 > 0,2/1

=> H2 dư, CuO hết, tính theo nCuO

=> n(H2,dư)=0,3-0,2=0,1(mol)

=> mH2(dư)=0,1.2=0,2(g)

a) \(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,4}{6}\) => Al hết, HCl dư

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

           0,1--->0,3------>0,1---->0,15

=> m_HCl = (0,4 - 0,3).36,5 = 3,65 (g)

b) V_H2 = 0,15.22,4 = 3,36 (l)

\(n_{Al}=\dfrac{m_{Al}}{M_{Al}}=\dfrac{2,7}{27}=0,1mol\)

\(n_{HCl}=\dfrac{m_{HCl}}{M_{HCl}}=\dfrac{14,6}{36,5}=0,4mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,1  <   0,4                                 ( mol )

0,1        0,3                         0,15 ( mol )

a. Chất còn dư là HCl

\(m_{HCl}=n_{HCl}.M_{HCl}=\left(0,4-0,3\right).36,5=3,65g\)

\(V_{H_2}=n_{H_2}.22,4=0,15.22,4=3,36l\)

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(n_{H2SO4}=\dfrac{14,7}{98}=0,15\left(mol\right)\)

PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

pư........0,15.......0,15..............0,15........0,15 (mol)

Ta có tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,15}{1}\) Vậy Mg dư, H₂SO₄ hết.

a) \(V_{H2}=22,4.0,15.\left(100\%-10\%\right)=3,024\left(l\right)\)

b) Chất dư sau pư là Mg

\(m_{Mg_{dư}}=24.\left(0,2-0,15\right)=1,2\left(g\right)\)

Vậy...........