a) Fe + H₂SO₄ --> FeSO₄ + H₂

b) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: Fe + H₂SO₄ --> FeSO₄ + H₂

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

=> V_H2 = 0,1.22,4 = 2,24 (l)

c) \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,1}{1}\) => CuO dư, H₂ hết

PTHH: CuO + H₂ --t^o--> Cu + H₂O

              0,1<--0,1------->0,1

=> m_CuO(dư) = (0,15 - 0,1).80 = 4 (g)

m_Cu = 0,1.64 = 6,4 (g)

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

b, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)

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

c, Ta có: \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)

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

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

Theo PT: \(n_{Cu}=n_{CuO\left(pư\right)}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow n_{CuO\left(dư\right)}=0,05\left(mol\right)\Rightarrow m_{CuO\left(dư\right)}=0,05.80=4\left(g\right)\)

\(m_{Cu}=0,1.64=6,4\left(g\right)\)

Bạn tham khảo nhé!

HD:

a,

2AL+3H2SO4=>AL2(SO4)3+3H2

b,

Ta có: nAL=10.8/27=0.4(mol)

theo phương trình ta có: nH2=3/2nAL=0.6(mol)

=> VCO2=0.6*22.4=13.44(lít)

c,

Ta có: nH2=11.2/22.4=0.5(mol)

theo phương trình ta có: nH2SO4=nH2=0.5(mol)

=>mH2SO4=0.5*98=49(g)

Fe + 2HCl -> FeCl2 + H2
nFe = 5,6/56 = 0,1 mol
=>nH2 = 0,1 mol
=> VH2= 0,1*22,4= 2,24 lít

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

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

=> \(\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,2.36,5}{15\%}=\dfrac{146}{3}\left(g\right)\\V_{H_2}=0,1.22,4=4,48\left(l\right)\end{matrix}\right.\)

a) 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

b) \(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

PTHH: 2Al + 3H₂SO₄ --> Al₂(SO₄)₃ + 3H₂

             0,4--->0,6-------------------->0,6

=> V_H2 = 0,6.22,4 = 13,44 (l)

c) \(V_{dd.H_2SO_4}=\dfrac{0,6}{1}=0,6\left(l\right)\)

d) \(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,6}{3}\) => Fe₂O₃ hết, H₂ dư

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

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

=> m_Fe = 0,2.56 = 11,2 (g)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\a, PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ b,n_{H_2}=n_{H_2SO_4}=n_{Fe}=0,2\left(mol\right)\\ V_{H_2\left(đkc\right)}=0,2.24,79=4,958\left(l\right)\\ c,C_{MddH_2SO_4}=\dfrac{0,2}{0,4}=0,5\left(M\right)\)

đổi 200 ml  = 0,02 l

a) PTHH : Fe + HCl -> FeCl₂ + H₂

b) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

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

\(C_{HCl}=\dfrac{n}{V}=\dfrac{0,1}{0,02}=5\left(M\right)\)

Đông Hải làm câu 1 rồi thì tui làm phần còn lại

Câu 2:

\(a,PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ b,n_{CuO}=\dfrac{m}{M}=\dfrac{8}{80}=0,1\left(mol\right)\\ Theo.PTHH:n_{Cu}=n_{H_2}=n_{CuO}=0,1\left(mol\right)\\ V_{H_2\left(đktc\right)}=n.22,4=0,1.22,4=2,24\left(l\right)\\ c,m_{Cu}=n.M=0,1.64=6,4\left(g\right)\\ d,PTHH:2H_2+O_2\underrightarrow{t^o}2H_2O\left(2\right)\\ Theo.PTHH\left(2\right):n_{H_2O}=n_{H_2}=0,1\left(mol\right)\)

\(m_{H_2O}=n.M=0,1.18=1,8\left(g\right)\)

a) \(n_{Fe}=\dfrac{1,12}{56}=0,02\left(mol\right)\)

PTHH: Fe + H₂SO₄ ---> FeSO₄ + H₂

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

b) \(V_{H_2}=0,02.22,4=0,448\left(l\right)\)

c) \(C\%_{H_2SO_4}=\dfrac{0,02.98}{100}.100\%=1,96\%\)

d) \(n_{CuO}=\dfrac{0,8}{80}=0,01\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

LTL: 0,01 < 0,02 => H₂ dư

Theo pthh: n_Cu = n_CuO = 0,01 (mol)

=> m_Cu = 0,01.64 = 0,64 (g)

a, PT: \(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

b, Ta có: \(n_{Fe}=\dfrac{33,6}{56}=0,6\left(mol\right)\)

Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,2\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4}=0,2.232=46,4\left(g\right)\)

c, Theo PT: \(n_{O_2}=\dfrac{2}{3}n_{Fe}=0,4\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,4.22,4=8,96\left(l\right)\) \(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{21\%}\approx42,67\left(l\right)\)

d, PT: \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)

Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{1}>\dfrac{0,4}{4}\), ta được Fe₃O₄ dư.

Theo PT: \(n_{Fe_3O_4\left(pư\right)}=\dfrac{1}{4}n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow n_{Fe_3O_4\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)

a. \(3Fe+2O_2\rightarrow Fe_3O_4\)

b. Số mol Fe: \(n=\dfrac{m}{M}=\dfrac{33,6}{56}=0,6\left(mol\right)\)

          PTHH: \(3Fe+2O_2\rightarrow Fe_3O_4\)

 Theo PTHH:  \(3\)        \(2\)            \(1\)           (mol)

       Theo đề:  \(0,6\)           \(\rightarrow0,2\)          (mol)

Kl của \(Fe_3O_4\) là: \(m=n\cdot M=0,2\cdot\left(56\cdot3+16\cdot4\right)=736\left(g\right)\)

\(n_{Fe}=\dfrac{m}{M}=\dfrac{33,6}{56}=0,6\left(mol\right)\\ PTHH:3Fe+2O_2-^{t^o}>Fe_3O_4\)

tỉ lệ:          3    :   2        :        1

n(mol)   0,6---->0,4-------->0,2

\(m_{Fe_3O_4}=n\cdot M=0,2\cdot\left(56\cdot3+16\cdot4\right)=46,4\left(g\right)\\ V_{O_2\left(dktc\right)}=n\cdot22,4=0,4\cdot22,4=8,96\left(l\right)\\ V_{kk}=8,96:21\%=42,6\left(l\right)\)

\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

\(PTHH:4H_2+Fe_3O_4-^{t^o}>3Fe+4H_2O\)

tỉ lệ:        4      :    1       :       3       :       4

n(mol)    0,4       0,2

n(mol pu) 0,4--->0,1----->0,3------->0,4

CL            0---->0,1------>0,3----->0,4

\(\dfrac{n_{H_2}}{4}< \dfrac{n_{Fe}}{1}\left(\dfrac{0,4}{4}< \dfrac{0,2}{1}\right)\)

=> `H_2` hết, `Fe` dư, tính theo `H_2`

\(m_{Fe_3O_4\left(dư\right)}=n\cdot M=0,1\cdot\left(56\cdot3+16\cdot4\right)=23,2\left(g\right)\)