a)
Fe_3O_4+4CO $$\xrightarrow{t^o}$$ 3Fe+4CO_2
BTKL: n_{O(o x i t)}=n_{CO}={11,24-9,64}/{16}=0,1(mol)
Theo PT: n_{Fe}=3/4n_{CO}=0,075(mol)
->m_{Fe}=0,075.56=4,2(g)
->m_{Cu}=9,64-4,2=5,44(g)
b)
n_{Fe_3O_4}=1/3n_{Fe}=0,025(mol)
n_{HCl}=0,2.1=0,2(mol)
Fe_3O_4+8HCl->2FeCl_3+FeCl_2+4H_2O
Do 0,025={0,2}/8-> Phản ứng hoàn toàn
->m_T=m_{Cu}=5,44(g)
Theo PT: n_{FeCl_3}=2n_{Fe_3O_4}=0,05(mol);n_{FeCl_2}=0,025(mol)
->m=0,05.162,5+0,025.127=11,3(g)