Bài 1 :

Ta có n_H2 = \(\dfrac{2,24}{22,4}\) = 0,1 ( mol )

R + 2HCl → RCl₂ + H₂

0,1..................0,1......0,1

=> 0,1 = \(\dfrac{5,6}{M_R}\)

=> M_R = 56

=> R là Fe

=> m_FeCl2 = 127 . 01 = 12,7 ( gam )

Ta có n_NaOH = 0,15 . 5,19 = 0,7785 ( mol )

CO₂ + 2NaOH → Na₂CO₃ + H₂O

x.............2x..............x................x
CO₂ + NaOH → NaHCO₃

y............y...................y

=> 2x + y = 0,7785

126x + 104y = 79,338

=> x = \(\dfrac{813}{41000}\)

y = 0,7388414634

=> n_CO2 = x + y = \(\dfrac{813}{41000}\) + 0,7388414634 = 0,7586292683 ( mol )

=> V_CO2 = 0,7586292683 . 22,4 = 16,99329561 ( lít )

=> m_Na2CO3 = \(\dfrac{813}{41000}\) . 126 = 2,498487805 ( gam )

=> m_NaHSO3 = 79,338 - 2,498487805 = 76,8395122 ( gam )