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Volume of co2 obtained at stp by the complete decomposition of 9.85 g na2co3 is.
Na2CO3⇒ Na2O + CO21 mole of sodium carbonate give 1 mole of carbon dioxide.as 105.98g sodium carbonate gives 44.01g carbon dioxidetherefore, 9.85 g sodium carbonate gives = 9.85 × 44.01 =4.09 g of CO2 105.98 Moles of CO2 mole = weight =4.09/44.01= 0.092 mole Molar mass1 mole of gas = 22.4 L gas Therefore,0.092 mole of CO2 = 22.4 × 0.092 = 2.06 L CO2.
Equation:Na2co3➡️Na2o+co2On calculating this reaction one might get answer 2.08 litres but as Na2co3 is soda ash and it does not decomposes on heating even to rednessAnd co2 does not evolved So the answer will be zero litres
Na2co3 requires temperature above 500 degree Celsius to decompose that is not possible at STP so the answer will be zero cow evolved
Na2CO3 on decomposition gives CO2 .This is unfair if , 9.85 gm of sodium carbonate decompose and 2.08 litres CO2 formed as per calculation. As it does not decompose formally by heating .
Na2CO3 require a high temperature(500℃).As the reaction happen in STP. So it can not get proper heat to make gasious CO2. So the volume of CO2 is Zero.....
It's answer will be zero bcz there will be no effect on Na2CO3 on decompositionIt need lots of energy to break the bonds ofNa2CO3 which is not possible in ordinary heating.
Hello Soumyadee,The solution of the problem is as followsNa2CO3⇒ Na2O + CO21 mole of sodium carbonate give 1 mole of carbon dioxideas 106 g sodium carbonate gives 44g carbon dioxidetherefore, 9.85 g sodium carbonate gives = 9.85 × 44 / 106 = 4.09 g of CO2Moles of CO2 mole =4.09/44 = 0.092 moleMolar mass 1 mole of gas = 22.4 L gasTherefore, 0.092 mole of CO2 = 22.4 × 0.092 = 2.06 L CO2.ThanksI hope above solution will clear your all doubts.Please feel free to post and ask as much doubts as possible.All the best.
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