a,
$C+O_2\buildrel{{t^o}}\over\to CO_2$
$C+CO_2\buildrel{{t^o}}\over\to 2CO$
$Fe_2O_3+3CO\buildrel{{t^o}}\over\to 2Fe+3CO_2$
b,
Trong $100kg$ gang:
$m_{Fe}=100.94\%=94kg$
Trong $Fe_2O_3$, $\%m_{Fe}=\dfrac{56.2.100}{160}=70\%$
$\Rightarrow m_{Fe_2O_3}=94:70\%=134,3kg$
$m_C=100.6\\%=6kg=m_{\text{than cốc}}$
c,
$m_{Fe}=94kg$
Sắt hao hụt $5\%$ nên cần điều chế $94+94.5\%=98,7kg$ sắt
$\Rightarrow m_{Fe_2O_3}=98,7:70\%=141kg$
$\Rightarrow m_{\text{quặng}}=141:90\%=156,67kg$