Bài 8 :
$a) n_{Fe} = \dfrac{16,8}{56} = 0,3(mol)$
$4Fe + 3O_2 \xrightarrow{t^o} 2Fe_2O_3$
$n_{O_2} = \dfrac{3}{4}n_{Fe} = 0,225(mol)$
$V_{O_2} = 0,225.22,4 = 5,04(lít)$
b)
$Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O$
$n_{Fe_2O_3} = \dfrac{1}{2}n_{Fe} = 0,15(mol)$
$n_{H_2SO_4} = 3n_{Fe_2O_3} = 0,45(mol)$
$m = 0,45.98 = 44,1(gam)$
Bài 9 :
5 tấn = 5000 kg
$n_{H_2O} = \dfrac{5000}{18} = \dfrac{2500}{9}(kmol)$
$6nCO_2 + 5nH_2O \xrightarrow{clorophin} (C_6H_{10}O_5)_n + 6nO_2$
$n_{tinh\ bột} = \dfrac{1}{5n}n_{H_2O} = \dfrac{500}{9n} (mol)$
$m_{tinh\ bột} = \dfrac{500}{9n}.162n = 9000(kg) = 9(tấn)$
