CBSE Class 11-science Answered

1. The reaction is given as:

Na₂CO₃ → Na₂O + CO₂ 

Therefore, 1 mole of Na₂CO₃ gives = 1 mole of CO₂ 
​
That is, 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 
​
Therefore,
Number 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 × 0.0929 L

                                        = 2.0809 L of CO_2​ would be obtained.