Question

Volume of CO2 obtained at STP by the complete decomposition of 9.85 g of Na2CO3is

Answer 1

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
​
Therefore
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.