Question

Answer 1

Caai 4 : 

a) $2A + 2H_2O \to 2AOH + H_2$

b) $n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)$

Theo PTHH : 

$n_A = 2n_{H_2} = 0,2(mol)$

Suy ra : $M_A = \dfrac{4,6}{0,2} = 23(Natri)$

 

Answer 2

Câu 5 :

Gọi $n_C = a(mol) ; n_S = b(mol)$

$\Rightarrow 12a + 32b = 2,2(1)$

$C + O_2 \xrightarrow{t^o} CO_2$

$S + O_2 \xrightarrow{t^o} SO_2$

$n_{O_2} = a + b = \dfrac{2,24}{22,4} = 0,1(2)$

Từ (1)(2) suy ra a = b = 0,05

$n_{CO_2} = n_C = 0,05(mol)$

$n_{SO_2} = n_S = 0,05(mol)$

Vậy :

$m_{CO_2} = 0,05.44 = 2,2(gam)$
$m_{SO_2} = 0,05.64 = 3,2(gam)$

Answer 3

Câu 6 :

a) $Fe + 2HCl \to FeCl_2 + H_2$

b)

Ta có : 

$m_{tăng} = m_{Fe} - m_{H_2}$

$⇔ 5,4 = 5,6 - m_{H_2}$

$⇔ m_{H_2} = 0,2(gam)$

$⇒ n_{H_2} = \dfrac{0,2}{2} = 0,1(mol)$
$⇒ V_{H_2} = 0,1.22,4 = 2,24(lít)$

c)

$n_{HCl} = 2n_{H_2} = 0,2(mol)$
$m_{HCl} = 0,2.36,5 = 7,3(gam)$