a, PT: \(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)

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

b, Theo PT: \(n_{H_2}=n_{H_2O}=a\left(mol\right)\)

Theo ĐLBT KL, có: m _oxit + m_H2 = m_KL + m_H2O

⇒ 24 + 2a = 17,6 + 18a ⇒ a = 0,4 (mol)

\(\Rightarrow V_{H_2}=0,4.22,4=8,96\left(l\right)\)

c, Gọi: \(\left\{{}\begin{matrix}n_{Fe_2O_3}=x\left(mol\right)\\n_{CuO}=y\left(mol\right)\end{matrix}\right.\) ⇒ 160x + 80y = 24 (1)

Theo PT: \(\left\{{}\begin{matrix}n_{Fe}=n_{Fe_2O_3}=2x\left(mol\right)\\n_{Cu}=n_{CuO}=y\left(mol\right)\end{matrix}\right.\) ⇒ 56.2x + 64y = 17,6 (2)

Từ (1) và (2) ⇒ x = y = 0,1 (mol)

\(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2O_3}=0,1.160=16\left(g\right)\\m_{CuO}=0,1.80=8\left(g\right)\end{matrix}\right.\)

d, \(\left\{{}\begin{matrix}\%m_{Fe_2O_3}=\dfrac{16}{24}.100\%\approx66,67\%\\\%m_{CuO}\approx33,33\%\end{matrix}\right.\)

CHÚC BẠN HỌC TỐT

a) Theo đề bài, ta có: \(n_{O2}=\dfrac{20}{32}=0,625\left(mol\right)\)

PTHH: \(2H_2+O_2\underrightarrow{o}2H_2O\)

pư............1.........0,5......1 (mol)

Ta có tỉ lệ: \(\dfrac{1}{2}< 0,625\). Vậy O2 dư, H2 hết.

\(\Rightarrow m_{H2O}=18.1=18\left(g\right)\)

Vậy.........

b) Đề nhầm bạn ơi, Cu chỉ có hóa trị là: I hoặc II

b) ... :) ?

\(n_{Cu}=\dfrac{6,4}{64}=0,1mol\)

\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)

0,1                           0,1            ( mol )

\(\left\{{}\begin{matrix}m_{CuO}=0,1.80=8g\\m_{FeO}=12-8=4g\end{matrix}\right.\)

PTHH: H2+FeO-----to-----> Fe+H2O

            H2+CuO------to---->Cu+H2O

a) 

FeO + H₂ --t^o--> Fe + H₂O

CuO + H₂ --t^o--> Cu + H₂O

b) \(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

              0,1<--0,1<-----0,1

=> \(m_{FeO}=12-0,1.80=4\left(g\right)\)

=> \(n_{FeO}=\dfrac{4}{72}=\dfrac{1}{18}\left(mol\right)\)

 FeO + H₂ --t^o--> Fe + H₂O

 \(\dfrac{1}{18}\)-->\(\dfrac{1}{18}\)----->\(\dfrac{1}{18}\)

=> \(V_{H_2}=\left(0,1+\dfrac{1}{18}\right).22,4=\dfrac{784}{225}\left(l\right)\)

c) \(m_{Fe}=\dfrac{1}{18}.56=\dfrac{28}{9}\left(g\right)\)

d) \(\left\{{}\begin{matrix}m_{CuO}=0,1.80=8\left(g\right)\\m_{FeO}=4\left(g\right)\end{matrix}\right.\)

\(a,\\ Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ Fe_3O_4+4H_2\rightarrow\left(t^o\right)3Fe+4H_2O\)

Loại phản ứng: Phản ứng thế

\(b,n_{Fe}=2.n_{Fe_2O_3}+3.n_{Fe_3O_4}=2.\dfrac{32}{160}+3.0,15=0,85\left(mol\right)\\ m_{Fe}=0,85.56=47,6\left(g\right)\\ c,n_{H_2}=\dfrac{32}{160}.3+4.0,15=1,2\left(mol\right)\\ V_{H_2\left(đktc\right)}=1,2.22,4=28\left(l\right)\)

Em xem sao oxit sắt lại hỏi KL nhôm nha! Vô lí!!!

\(n_{CuO}=a\left(mol\right),n_{Fe_2O_3}=b\left(mol\right)\)

\(m=80a+160b=6\left(g\right)\left(1\right)\)

\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

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

\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}2Fe+3H_2O\)

\(n_{H_2}=a+3b=0.1\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.025,b=0.025\)

\(m_{kl}=0.025\cdot64+0.025\cdot2\cdot56=4.4\left(g\right)\)

\(b.\)

\(m_{hh}=3m_{Fe_2O_3}=6\left(g\right)\)

\(\Rightarrow n_{Fe_2O_3}=\dfrac{2}{160}=0.0125\left(mol\right)\)

\(\Rightarrow n_{CuO}=0.0125\left(mol\right)\)

\(m_{kl}=0.0125\cdot2\cdot56+0.0125\cdot64=2.2\left(g\right)\)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\\ Theo.pt:n_{Fe}=n_{H_2}=0,3\left(mol\right)\\ m_{Fe}=0,3.56=16,8\left(g\right)\\ m_{Cu}=20-16,8=3,2\left(g\right)\\ n_{Cu}=\dfrac{3,2}{64}=0,06\left(mol\right)\\ PTHH:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ Mol:0,05\leftarrow0,05\leftarrow0,05\\ m_{CuO}=0,05.80=4\left(g\right)\)

a.\(n_{H_2}=\dfrac{6,72}{22,4}=0,3mol\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,3                                   0,3 ( mol )

\(m_{Fe}=0,3.56=16,8g\)

\(\%m_{Fe}=\dfrac{16,8}{20}.100=84\%\)

\(\%m_{Cu}=100\%-84\%=16\%\)

b.\(m_{Cu}=20-16,8=3,2g\)

\(n_{Cu}=\dfrac{3,2}{64}=0,05mol\)

\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)

0,05                      0,05             ( mol )

\(m_{CuO}=0,05.80=4g\)

a) Ta giải hệ: \(m_{CuO}+m_{Fe_2O_3}=37,6\) và \(m_{CuO}-m_{Fe_2O_3}=-15,2\)