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

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

Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.22,4=3,36\left(l\right)\)

b, Có lẽ đề hỏi bao nhiêu gam đồng thay vì "bao nhiêu gam sắt" bạn nhỉ?

\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)

PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

Xét tỉ lệ: \(\dfrac{0,25}{1}>\dfrac{0,15}{1}\), ta được CuO dư.

Theo PT: \(n_{Cu}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Cu}=0,15.64=9,6\left(g\right)\)

a) nAl=2,7/27=0,1(mol)

nHCl=14,6/36,5= 0,4(mol)

PTHH: 2Al +6 HCl -> 2 AlCl3 +3 H2

Ta có: 0,1/2 < 0,6/4

=> HCl dư, Al hết, tính theo nAl

=> nAlCl3=nAl=0,1(mol)

=> mAlCl3=0,1.133,5=13,35(g)

b) nH2= 3/2. nAl=3/2. 0,1=0,15(mol)

=>V(H2,đktc)=0,15.22,4=3,36(l)

c) mFe2O3(nguyên chất)= 80%. 38,4=30,72(g)

=>nFe2O3= 30,72/160=0,192(mol)

PTHH: Fe2O3 + 3 H2 -to->2 Fe +3 H2O

Ta có: 0,192/1 > 0,15/3

=> H2 hết, Fe2O3 dư, tính theo nH2

=> nFe= 2/3. nH2= 2/3. 0,15=0,1(mol)

=>mFe=0,1.56=5,6(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 

Mol:     0,1                    0,1        0,15

Tỉ lệ:\(\dfrac{0,1}{2}< \dfrac{0,4}{6}\) ⇒ Al pứ hết,HCl dư

\(\Rightarrow m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\)

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

c,\(m_{Fe_2O_3\left(tinhkhiét\right)}=38,4.\left(100\%-20\%\right)=30,72\left(g\right)\)

⇒\(n_{Fe_2O_3}=\dfrac{30,72}{160}=0,192\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ → 2Fe + 3H₂O 

Mol :                  0,15       0,1

Tỉ lệ:\(\dfrac{0,192}{1}>\dfrac{0,15}{3}\)⇒ Fe₂O₃ dư,H₂ hết

=> m_Fe = 0,1.56 =5,6 (g)

a, \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

b, Ta có: \(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

\(m_{Cu}=6-2,8=3,2\left(g\right)\Rightarrow n_{Cu}=\dfrac{3,2}{64}=0,05\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Cu}+\dfrac{3}{2}n_{Fe}=0,125\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,125.22,4=2,8\left(l\right)\)

\(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right);n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\\ PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,4}{1}>\dfrac{0,25}{1}\Rightarrow Fe.dư\\ n_{H_2}=n_{Fe\left(p.ứ\right)}=n_{H_2SO_4}=0,25\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\\ b,n_{Fe\left(dư\right)}=0,4-0,25=0,15\left(g\right)\\ m_{Fe\left(dư\right)}=0,14.56=8,4\left(g\right)\)

a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

           0,1-------------->0,1------>0,1

\(\Rightarrow m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)

b) \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c) \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

     0,1<---0,1

\(\Rightarrow m_{CuO}=0,1.80=8\left(g\right)\)

a: Zn+2HCl->ZnCl2+H2

0,2       0,4      0,2       0,2

mZnCl2=0,2*136=27,2(g)

b: V=0,2*22,4=4,48(lít)

n_mg = \(\dfrac{3,6}{24}\)  = 0,15 ( mol )
Mg +  2HCl → MgCl₂ + H_{2
1          2            1           1
0,15      0,3                    0,15}
a) m_HCl = 0,3 . 36,5 = 10,95 ( g )
b) V_H2 = 0,15 . 22,4 = 3,36 ( l )
Tick dùm tớ nha, tớ giải theo trường có gì sai bỏ qua nhaaa

Giải thích các bước giải:

 a) Mg +H2SO4--->MGSO4+H2

n Mg =6/24=0,25(mol)

n H2=n Mg =0,25(mol)

V H2=0,25.22,4=5,6(l)

b) 3H2+FE2O3-->2Fe+3H2O

n Fe2O3=32/160=0,2(mol)

->Fe2O3 dư

n Fe =2/3n H2=1/6(mol)

m Fe =1/6.56=28/3(g)

\(a,n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ n_{H_2}=n_{Mg}=0,25\left(mol\right)\\ V_{H_2\left(\text{đ}ktc\right)}=0,25.22,4=5,6\left(l\right)\\ b,n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\\ 3H_2+Fe_2O_3\rightarrow\left(t^o\right)2Fe+3H_2O\\ n_{Fe}=2.n_{Fe_2O_3}=2.0,2=0,4\left(mol\right)\\ m_{Fe}=0,4.56=22,4\left(g\right)\)

nZn= 19,5/65=0,3(mol); nFe2O3=19,2/160=0,12(mol)

PTHH: Zn + 2 HCl -> ZnCl2 + H2

Fe2O3 + 3 H2 -to-> 2 Fe +3 H2O

nH2=nZnCl2= nZn=0,3(mol) => V(H2,đktc)=0,3.22,4= 6,72(l)

b) nHCl= 2.0,3=0,6(mol) => mHCl=0,6.36,5=21,9(g)

=>mddHCl=(21,9.100)/20=109,5(g)

=>m=109,5(g)

c) mH2=0,3.2=0,6(mol)

mddZnCl2=19,5+109,5 - 0,6= 128,4(g)

mZnCl2=0,3. 136= 40,8(g)

=>C%ddZnCl2= (40,8/128,4).100=31,776%

d) Ta có: 0,3/3 < 0,12/1

=> H2 hết, Fe2O3 dư, tính theo nH2

=> nFe= 2/3. nH2= 2/3. 0,3= 0,2(mol)

=>mFe=0,2.56=11,2(g)

a, n_Zn = 19,5/65=0,3 (mol)

PTHH: Zn + 2HCl → ZnCl₂ + H₂ 

Mol:    0,3     0,15        0,3      0,3

=> \(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

b,m_HCl=0,15.36,5=5,475 (g)

=> m=m_ddHCl=5,475:20%=27,375 (g)

c,m_{dd sau pứ} =19,5+27,375=46,875 (g)

\(m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{40,8}{46,875}.100\%=87,04\%\)

d,\(n_{Fe_2O_3}=\dfrac{19,2}{160}=0,12\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ → 2Fe + 3H₂O

Mol:                     0,3       0,2    

 Tỉ lệ: 0,12/1>0,3/3 ⇒ Fe₂O₃ dư,H₂ pứ hết

=> m_Fe=0,2.56=11,2 (g)