a, \(n_{CO}=\dfrac{22,4}{22,4}=1\left(mol\right);n_{Fe}=\dfrac{16}{160}=0,1\left(mol\right)\)

PTHH: Fe₂O₃ + 3CO ---t^o→ 2Fe + 3CO₂

Mol:       0,1          0,3             0,2

Ta có: \(\dfrac{0,1}{1}< \dfrac{1}{3}\) ⇒ Fe hết, CO dư

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

b, \(m_{Fe}=0,2.56=11,2\left(l\right)\)

a. Công thức về khối lượng:

\(m_{Fe_2O_3}+m_{H_2}=m_{Fe}+m_{H_2O}\)

b. Áp dụng câu a, ta có:

\(m_{Fe_2O_3}+2=56+18\)

\(\Leftrightarrow m_{Fe_2O_3}=56+18-2\)

\(\Leftrightarrow m_{Fe_2O_3}=72\left(g\right)\)

\(a)3H_2+Fe_2O_3-^{t^o}\rightarrow2Fe+3H_2O\\b)BTKL:m_{H_2}+m_{Fe_2O_3}=m_{Fe}+m_{H_2O}\\ \Leftrightarrow2+m_{Fe_2O_3}=56+18 \\ \Rightarrow m_{Fe_2O_3}=72\left(g\right)\)

Bn phải ghi rõ là oxit nào nha.

a. PT: Fe₂O₃ + 3CO ---> 2Fe + 3CO₂.

b. Ta có: \(n_{Fe_2O_3}=\dfrac{32}{160}=0,2\left(mol\right)\)

n_CO = \(\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

Ta thấy: \(\dfrac{0,2}{1}>\dfrac{0,3}{3}\)

Vậy Fe dư.

c. Theo PT: n_Fe = 2.n_CO = 2 . 0,3 = 0,6(mol)

=> m_Fe = 0,6 . 56 = 33,6(g)

Theo PT: \(n_{CO_2}=n_{CO}=0,3\left(mol\right)\)

=> \(m_{CO_2}=0,3.44=13,2\left(g\right)\)

\(n_K=\dfrac{3.9}{39}=0.1\left(mol\right)\)

\(K+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\)

\(0.1..................0.1......0.05\)

\(m_{KOH}=0.1\cdot56=5.6\left(g\right)\)

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

\(n_{CuO}=\dfrac{20}{80}=0.25\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)

\(1.........1\)

\(0.25.......0.05\)

\(LTL:\dfrac{0.25}{1}>\dfrac{0.05}{1}\Rightarrow CuOdư\)

\(m_Z=m_{Cu}+m_{CuO\left(dư\right)}=0.05\cdot64+\left(0.25-0.05\right)\cdot80=19.2\left(g\right)\)

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

a) Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\\n_{Fe_2O_3}=\dfrac{24}{160}=0,15\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,15}{1}< \dfrac{0,5}{3}\) \(\Rightarrow\) Fe₂O₃ p/ứ hết, H₂ còn dư

\(\Rightarrow n_{H_2\left(dư\right)}=0,05\left(mol\right)\)

b) 

+) Cách 1

Theo PTHH: \(n_{Fe}=2n_{Fe_2O_3}=0,3\left(mol\right)\) \(\Rightarrow m_{Fe}=0,3\cdot56=16,8\left(g\right)\)

+) Cách 2:

Bảo toàn nguyên tố: \(n_{Fe}=2n_{Fe_2O_3}=....\)

\(a,n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\\ PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\\ Mol:0,05\rightarrow0,15\rightarrow0,1\\ m_{Fe}=0,1.56=5,6\left(g\right)\\ b,V_{H_2}=0,15.22,4=3,36\left(l\right)\\ c,n_{O_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ \\ PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ LTL:\dfrac{0,1}{3}>\dfrac{0,05}{2}\Rightarrow Fe.dư\\ n_{Fe_3O_4}=\dfrac{0,05}{2}=0,025\left(mol\right)\\ m_{Fe_3O_4}=0,025.232=5,8\left(g\right)\)

nFe2O3 = 8 : 160 = 0,05 (mol) 
pthh: Fe2O3 + 3H2 -t--> 2Fe + 3H2O
         0,05--------0,15----->0,1 (mol) 
=> VH2= 0,15 . 22,4 = 3,36 (L) 
=> mFe = 0,1 . 56 = 5,6 (g) 
nO2  = 1,12 : 22,4 = 0,05 (mol) 
pthh : 2H2+ O2 -t-> 2H2O
 LTL :
0,15/2   > 0,05/1
=> H2 du 
theo pt , nH2O = 2 nO2 = 0,1 (mol) 
=> mH2O = 0,1 .18 = 1,8 (g) 

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

0,2----------0,6------0,4-----0,6 mol

n H2O=\(\dfrac{10,8}{18}\)=0,6 mol

=>VH2=0,6.22,4=13,44l

b)m Fe=0,4.56=22,4g

c) m Fe2O3=0,2.160=32g

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

a+b) \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2O_3}=0,2\left(mol\right)\\n_{H_2}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2O_3}=0,2\cdot160=32\left(g\right)\\V_{H_2}=0,6\cdot22,4=13,44\left(l\right)\end{matrix}\right.\)

c) PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

Theo PTHH: \(n_{Zn}=n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,6\cdot65=39\left(g\right)\)

a,

nFe = 22,4/56 = 0,4  (mol)

PTHH

Fe₂O₃ + 3H₂ ---t^o----) 2Fe + 3H₂O        (1)

theo phương trình (1) ,ta có:

nFe₂O₃ = 0,4 x 2 / 1 = 0,8 (mol)

mFe₂O₃ = 160 x 0,8 = 128 (g)

b,

theo pt (1)

nH₂ = (0,4 x 3)/2 = 0,6 (mol)

=) VH₂ = 0,6 x 22,4 = 13,44 (L)

c,

PTHH

Zn + H₂SO₄ -------------) ZnSO₄ + H₂                         (2)

Số mol H₂ cần dùng là 0,6 (mol)

Theo PT (2) :

nZn = nH₂    ==) nZn = 0,6 x 65 = 39 (g)