Question

Answer 1

Câu 4:

Sau p/ứ nung nóng, ta thấy CO còn dư, oxit p/ứ hết

a) Ta có: \(\left\{{}\begin{matrix}n_{hh\left(khí\right)}=\dfrac{2,688}{22,4}=0,12\left(mol\right)\\n_{O\left(oxit\right)}=\dfrac{m_{oxit}-m_{KL}}{16}=\dfrac{6,22-4,94}{16}=0,08\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,04\left(mol\right)=n_{CO_2}\\n_{CO}=0,12-0,04=0,08\left(mol\right)\end{matrix}\right.\) \(\Rightarrow d_{D/H_2}=\dfrac{0,04\cdot44+0,08\cdot28}{2}=2\)

b) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)

Ta có: \(n_{H_2}=\dfrac{1,568}{22,4}=0,07\left(mol\right)=\Sigma n_{Fe}\)

PTHH: \(Fe_3O_4+4CO\xrightarrow[]{t^o}3Fe+4CO_2\)

Ta có: \(n_{Fe_3O_4}=\dfrac{1}{4}n_{CO_2\left(p/ứ\right)}=0,01\left(mol\right)\) \(\Rightarrow m_{Fe_3O_4}=0,01\cdot252=2,52\left(g\right)\)

Mặt khác: \(n_{Fe}=0,07-0,01\cdot3=0,04\left(mol\right)\) \(\Rightarrow m_{Fe}=0,04\cdot56=2,24\left(g\right)\)

\(\Rightarrow m_{Al_2O_3}=6,22-2,52-2,24=1,46\left(g\right)\)