Đáp án:

Giải thích các bước giải:

 Cốc `1 : `

`n_(Na_2CO_3) =16/106 = 8/53 `

`Na_2CO_3 + 2HCl -> 2NaCl + CO_2 + H_2O`

`8/53 -> 16/53 -> 16/53 -> 8/53 ` 

`m_(1) = 16 - 8/53 * 44 = 496/53 `

Cốc `2 : `

`2Al + 3H_2SO_4 -> Al_2(SO_4)_3 + 3H_2` 

`a -> 1.5a -> 0.5a -> 1.5a `

Ta có : `m_(1) = m_(2) ` 

`-> 27a - 1.5a * 2= 496/53 ` 

`-> x = 62/159 `

`-> m =27 * 62/159 = 10.53 (gam)` 

· Cốc A:

$Na_2CO_3+2HCl\to 2NaCl+CO_2+H_2O\\ n_{CO_2}=n_{Na_2CO_3}=\dfrac{16}{106}=\dfrac{8}{53}(mol)\\ m_{A\ tang}=m_{Na_2CO_3}-m_{CO_2}\\ =16-\dfrac{8}{53}.44=\dfrac{496}{53}(g)$

· Cốc B:

$2Al+3H_2SO_4\to Al_2(SO_4)_3-3H_2\\ n_{Al(pu)}:a(mol)\\ n_{H_2}=\dfrac{3}{2}n_{Al}=1,5a(mol)\\ m_{B\ tang}=m_{Al}-m_{H_2}\\ =27a-1,5a.2=24a(g)$

· Để cân thăng bằng:

$m_{A\ tang}=m_{B\ tang}\\ hay\  \dfrac{496}{53}=24a\\ \to a=\dfrac{62}{159}(mol)\\ m_{Al}=\dfrac{62}{159}.27=10,528(g)$