a,
$m_{Fe_2O_3}=160.80\%=128g$
$\to m_{Fe_2O_3\text{pứ}}=128.75\%=96g$
$\to m_{Fe_2O_3\text{dư}}+m_{\text{tạp chất}}=160-96=64g$
$n_{Fe_2O_3\text{pứ}}=\dfrac{96}{160}=0,6(mol)$
$Fe_2O_3+3CO\xrightarrow{{t^o}} 2Fe+3CO_2$
Theo PTHH:
$n_{CO_2}=3n_{Fe_2O_3}=0,6.3=1,8(mol)$
$\to V_B=1,8.22,4=40,32l$
b,
Theo PTHH:
$n_{Fe}=2n_{Fe_2O_3}=1,2(mol)$
Vậy $m_A=m_{Fe_2O_3\text{dư}}+m_{\text{tạp chất}}+m_{Fe}=64+1,2.56=131,2g$