Question

Answer 1

Bài 1 : 

$Fe + 2HCl \to FeCl_2 + H_2$
$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

Theo PTHH : 

$n_{CuO} =  n_{FeCl_2} = n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$m = 0,2.80 = 16(gam)$

$m_{FeCl_2} = 0,2.127 = 25,4(gam)$

Answer 2

a) $n_{Fe} = 0,3(mol)$
$4F e+ 3O_2 \xrightarrow{t^o} 2Fe_2O_3$

Theo PTHH :

$n_{O_2} = \dfrac{3}{4}n_{Fe} = 0,225(mol)$
$V_{O_2} = 0,225.22,4 =5,04(lít)$
b)

$n_{Fe_2O_3} = \dfrac{1}{2}n_{Fe} = 0,15(mol)$

$Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O$
$n_{H_2SO_4} = 3n_{Fe_2O_3} = 0,45(mol)$
$m = 0,45.98 = 44,1(gam)$