$2)$

$Fe_2O_3(s)+3CO(g) \xrightarrow{t^o}3Fe(g)+3CO_2(g)$

$\Delta _r H^o_{298K}=(-393.5) \times 3-(-110.5) \times 3-(-824.2)=-24.8(kJ/mol)$

$n_{Fe_2O_3}=0.01(kmol)$

$n_{CO}=0.025(kmol)$

$\dfrac{0.01}{1}>\dfrac{0.025}{3}$

`=>`$Fe_2O_3$ dư tính theo $CO.$

`=>`$n_{Fe_2O_3,pu}=\dfrac{1}{3}n_{CO}=\dfrac{1}{120}(kmol)$

`=>` $\Delta _rH^o_{298K}=\dfrac{1}{120} \times 10^3 \times (-24.8)=-206.67(kJ)$

Đáp án:

$\rm n_{Fe_2O_3}=\dfrac{1,6}{160}=0,01\ (kmol)\\n_{CO}=\dfrac{0,7}{12}=0,025\ (mol)$

Phản ứng xảy ra:

$\rm Fe_2O_3+3CO\xrightarrow{t^°}2Fe+3CO_2$

Tỉ lệ: $\rm\dfrac{n_{Fe_2O_3}}{1}>\dfrac{n_{CO}}{3}\to Fe_2O_3$ dư, tính theo $\rm CO$.

$\rm n_{Fe_2O_3\ (pu)}=\dfrac{1}{3}n_{CO}=\dfrac{1}{200}\ (kmol)$

Biến thiên enthalpy khi 1 kmol $\rm Fe_2O_3$ phản ứng:

$\rm \Delta _r H^°_{298\ (1)}=3\Delta_fH^°_{298\ (CO_2)}-\Delta_fH^°_{298\ (Fe_2O_3)}-\Delta_fH^°_{298\ (CO)}\\\kern2cm=(-393,5) \times 3- \times 3-(-824,2)-(-110,5)\\\kern2cm=-24,8\ (kJ)$

Biến thiên enthalpy khi $\dfrac{1}{200}$ kmol $\rm Fe_2O_3$ phản ứng:

$\rm \Delta _r H^°_{298\ (2)}=n_{Fe_2O_3\ (pu)}×\Delta _r H^°_{298\ (1)}\kern2cm=\dfrac{1}{120}×10^3×(-24,8)=-206,67\ (kJ)$