Question

The volume of CO2 obtained by complete decomposition of one mole of NaHCO3 at STP is

Answer 1

Na2CO3 ⇒  Na2O + CO2

1 mole of sodium carbonate give 1 mole of carbon dioxide.

      105.98 g [M.W.] sodium carbonate gives 44.01 g [M.W.] carbon dioxide

     ∴  9.85 g sodium carbonate gives = ?                                                                                                              = 9.85 × 44.01                                                                                                       105.98                                                                          = 4.09 g of CO2

Moles of CO2 mole = weight     

                                Molar mass 

                             = 4.09    

                                44.01       

                            = 0. 092 mole

1 mole of gas = 22.4 L gas  

∴ 0.092 mole of CO2  = 22.4 × 0.092 = 2.06 L CO2 obtain  

Answer 2

Answer:

The volume of CO2 obtained by complete decomposition of one mole of NaHCO3 at STP is 22.4 L.

Explanation:

To calculate the number of moles of sodium carbonate at STP:

2$NaHCO_{3}$ ⇒ $Na_{2}$$CO_{3}$ + $CO_{2}$ + $H_{2}$$O_{}$...................................(i)

• From above equation , we can see that 2 mole of NaHCO3 is decomposed into 1 mole of Na2CO3, 1 mole of CO2 and 1 mole of water (H2O).

• At STP(Standard Temperature and Pressure ) :

    1 mole of a gas occupies 22.4 L of volume.

    Hence, 2 moles of NaHCO3 can produce 1 mole of CO2 or 22.4 L of         CO2.

• 1 moles of NaHCO3 can produce 22.4/2 L = 11.2 L of CO2.

• The volume of CO2 obtained by complete decomposition of one mole of NaHCO3 at STP is 22.4 L.