Question

Vol of CO2 obtained at STP by the complete decomposition of 9.85 g of Na2CO3 will be ?

Answer 1

⭕️reaction will be :-

Na2CO3 → Na2O+CO2

⭕️Thus 1 mole of Na2CO3 gives =1 mole of CO2

​i.e

⭕️105.98 g of Na2CO3 gives = 44g of CO2

⭕️​9.85 g of Na2CO3​ gives = 44105.98×9.85 g
 
⇒4.0894 g of CO2

​∴

⭕️No. of moles of CO2 produced  = mass/molar mass =4.0894/44
= 0.0929 moles

⭕️1 mole of CO2 gas =22.4 L

0.0929 moles of gas At STP = 22.4 x 0.0929

⇒ ️∴ L= 2.0809 L of CO2​ would be obtained.

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Answer 2

Answer:

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