Question

[huy]

giúp mình nhanh nhé!! mình camon:>

Answer 1

[Nguyễn Ngọc Nam]

a. PTHH: \(Fe_3O_4+4CO\rightarrow^{t^o}3Fe+4CO_2\uparrow\)

Đặt \(x\left(mol\right)=n_{Fe_3O_4\left(\text{p/ứ}\right)}\)

\(\rightarrow m_{Fe_3O_4}\left(\text{p/ứ}\right)=232x\left(g\right)\)

Theo phương trình \(n_{Fe}=3n_{Fe_3O_4}=3x\left(mol\right)\)

\(\rightarrow m_{Fe_3O_4\left(\text{dư}\right)}=62-3x.56=62-168x\left(g\right)\)

Có \(m_{Fe_3O_4\left(\text{p/ứ}\right)}+m_{Fe_3O_4\left(\text{dư}\right)}=81,2g\)

\(\rightarrow232x+62-168x=81,2\)

\(\rightarrow x=0,3mol\)

\(\rightarrow m_{Fe_3O_4\left(\text{p/ứ}\right)}=0,3.232=69,6g\)

b. Theo phương trình \(n_{CO}=4n_{Fe_3O_4\left(\text{p/ứ}\right)}=1,2mol\)

\(\rightarrow V_{CO\left(ĐKTC\right)}=1,2.22,4=26,88l\)

c. \(m_{Fe_3O_4\left(\text{dư}\right)}=62-168.0,3=11,6g\)

\(H\%=\frac{69,6}{69,6+11,6}.100\%\approx85,71\%\)