Answers

2016-03-03T19:08:02+05:30
Solution....


1 mole of Na2CO3 gives 1 mole of CO2

105.98g of Na2CO3 = 44g of CO2
therefore,
            9.85
g of Na2CO3 = (44*9.85)/105.98 g of CO2
                                         =4.0894 g
Number of moles of CO2 = mass / molar mass
                                         = 44/
4.0894 = 0.0929 moles

1 mole of CO2 = 22.4 l

0.0929 moles = 22.4*0.0929 = 2.0809 l

Volume of CO2 = 2.0809 l








Equation is given in the pic below...


0
sorry pic not sent..
Na2CO3 ------> Na2O + CO2