a)

Gọi \(n_{H_2O} = n_{H_2\ pư} = a(mol)\)

Bảo toàn khối lượng :

\(m_{Fe_2O_3} + m_{H_2\ pư} = m_A + m_{H_2O}\\ \Leftrightarrow 32 + 2a = 18a + 24,8\\ \Leftrightarrow a = 0,45(mol)\\ \Rightarrow V_{H_2} = 0,45.22,4 = 10,08(lít)\)

b)

\(Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\\ n_{Fe} = \dfrac{2}{3}n_{H_2} = 0,3(mol)\\ \Rightarrow \%m_{Fe} = \dfrac{0,3.56}{24,8}.100\% = 67,74\%\\ \%m_{Fe_2O_3\ dư} = 100\% - 67,74\% = 32,26\% \)

c)

\(m_{Fe_2O_3\ dư} = 24,8 - 0,3.56 = 8(gam)\\ \Rightarrow H = \dfrac{32-8}{32}.100\% = 75\% \) 

PT: \(CuO+H_2\rightarrow Cu+H_2O\)

Gọi \(n_{H_2}=x\left(mol\right)\)

Theo PT: \(n_{H_2O}=n_{H_2}=x\left(mol\right)\)

Theo ĐLBT KL, có: m_CuO + m_H2 = m _{chất rắn} + m_H2O

⇒ 12 + 2x = 10,4 + 18x ⇒ x = 0,1 (mol)

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

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

PTHH : \(Fe_2O_3+3H_2-t^o->2Fe+3H_2O\)

=> Chất rắn A : Fe

a)  \(nFe_2O_3=\frac{32}{160}=0,2\left(n\right)\); \(n_{Fe}=\frac{24,8}{56}=0,4\left(n\right)\)

=> \(\frac{0,2}{2}< \frac{0,4}{2}\)=> Fe dư

Theo PT : \(nH_2=3.nFe_2O_3=0,2\times3=0,6\left(n\right)\)

=> \(V_{H_2\left(đktc\right)}=0,6\times22,4=13,44\left(l\right)\)

wtf, nani

Bùi Quang Nguyên làm dc ko?

\(n_{Cu}=\dfrac{3.2}{64}=0.05\left(mol\right)\)

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

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

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

\(m_{CuO}=0.05\cdot80=4\left(g\right)\)

a) PTHH : \(3H_2+Fe_2O_3-t^o->2Fe+3H_2O\)

Ta có : \(m_{CR\left(giảm\right)}=m_{O\left(lay.di\right)}\)

=> \(m_{O\left(lay.di\right)}=32-24,8=7,2\left(g\right)\)

=> \(n_{O\left(lay.di\right)}=\frac{7,2}{16}=0,45\left(mol\right)\)

Theo pthh : \(n_{H_2\left(pứ\right)}=n_{O\left(lay.di\right)}=0,45\left(mol\right)\)

=> \(V_{H_2\left(pứ\right)}=0,45\cdot22,4=10,08\left(l\right)\)

b) Theo pthh : \(n_{Fe\left(spu\right)}=\frac{2}{3}n_{H_2\left(pứ\right)}=0,3\left(mol\right)\)

=> \(m_{Fe}=16,8\left(g\right)\)

=> \(\hept{\begin{cases}\%m_{Fe}=\frac{16,8}{24,8}\cdot100\%\approx67,74\%\\\%m_{Fe_2O_3}\approx100\%-67,74\%=32,26\%\end{cases}}\)

c) Theo pthh : \(n_{Fe_2O_3\left(bi.khu\right)}=\frac{1}{3}n_{H_2\left(pứ\right)}=0,15\left(mol\right)\)

Mà thực tế, \(n_{Fe_2O_3}=\frac{32}{160}=0,2\left(mol\right)\)

=> \(H\%=\frac{0,15}{0,2}\cdot100\%=75\%\)

Tham khảo:

Gọi nCuO phản ứng là a (mol)

PTHH: CuO + H2O -> (t°) Cu + H2O

Mol: a ---> a ---> a ---> a

nCuO (bđ) = 24/80 = 0,3 (mol)

Ta có: 80 . (0,3 - a) + 64a = 22,4

<=> a = 0,1 (mol)

%mCu = (0,1 . 64)/22,4 = 28,57%

%mCuO = 100% - 28,57% = 71,43%

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

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)