\(n_{Fe_2O_3}=\dfrac{12}{160}=0,075\left(mol\right)\\a, Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\\ b,n_{H_2}=3.0,075=0,225\left(mol\right)\\ V_{H_2\left(đkc\right)}=24,79.0,225=5,57775\left(l\right)\\ c,n_{Fe}=2.0,075=0,15\left(mol\right)\\ m_{Fe}=0,15.56=8,4\left(g\right)\)

PTHH: \(Fe_2O_3+3CO\xrightarrow[]{t^o}2Fe+3CO_2\uparrow\)

Đổi 1 tấn = 1000 kg

Ta có: \(n_{Fe}=\dfrac{1000\cdot98\%}{56}=17,5\left(kmol\right)\) 

\(\Rightarrow n_{CO}=26,25\left(kmol\right)\) \(\Rightarrow V_{CO}=26,25\cdot22,4=588\left(m^3\right)\)

1) PTHH: 2Fe₂O₃ + 3CO →4Fe + 3CO₂

2) n_co=\(\dfrac{V}{22,4}\)=\(\dfrac{3,36}{22,4}=0,15\)(mol)

-Theo PTHH, ta có:

2.n_Fe2O3=3.n_CO=4.n_Fe=3.n_CO2=3.0,15=0,45(mol)

=>n_Fe2O3=\(\dfrac{0,45}{2}=0,225\left(mol\right)\)

=>m_Fe2O3=n.M=0,225.(56.2+16.3)=36(g)

c)- Ta có: 3.n_CO2=3.0,15=0,45(mol)

=>n_CO2=\(\dfrac{0,45}{3}=0,15\left(mol\right)\)

=>V_CO2=n.22,4=0,15.22,4=3,36(lít)

áp dụng định luật bảo toàn khối lượng, ta có:

\(m_{CO}+m_{Fe_2O_3}=m_{Fe}+m_{CO_2}\)

\(\Rightarrow m_{CO_2}=m_{CO}+m_{Fe_2O_3}-m_{Fe}\)

              \(=12,4+18,6-15,2=15,8 \left(g\right)\)

vậy khối lượng khí \(CO_2\) thu được là  \(15,8g\)

\(n_{CO}=a\left(mol\right)\)

\(\Rightarrow n_{CO_2}=a\left(mol\right)\)

Áp dụng định luật bảo toàn khối lượng :

\(m_{Fe_2O_3}+m_{CO}=m_{cr}+m_{CO_2}\)

\(\Rightarrow48+28a=43.2+44a\)

\(\Rightarrow a=0.3\)

\(V_{CO}=0.3\cdot22.4=6.72\left(l\right)\)

Đáp án A

\(n_{Fe_2O_3}=\dfrac{11,2}{160}=0,07mol\\ Fe_2O_3+3CO\xrightarrow[t^0]{}2Fe+3CO_2\\ n_{Fe}=2.0,07=0,14mol\\ m_{Fe}=0,14.56=7,84g\)

\(n_{Fe_2O_3}=\dfrac{11,2}{160}=0,07\left(mol\right)\)

PTHH :

\(Fe_2O_3+3CO\rightarrow2Fe+3CO_2\)

0,07                      0,14

\(m_{Fe}=0,14.56=7,84\left(g\right)\)

Ta có: \(n_{Fe_2O_3}=\dfrac{11,2}{160}=0,07\left(mol\right)\)

PT: \(Fe_2O_3+3CO\underrightarrow{t^o}2Fe+3CO_2\)

Theo PT: \(n_{Fe}=2n_{Fe_2O_3}=0,14\left(mol\right)\)

\(\Rightarrow m_{Fe}=0,14.56=7,84\left(g\right)\)