Volume of carbon dioxide at STP by complete decomposition of 9.85g of sodium carbonate
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Na2CO3 ⇒ Na2O + CO2
1 mole of sodium carbonate give 1 mole of carbon dioxide.
105.98 g of sodium carbonate gives 44 g of carbon dioxide.
Then, 9.85 g of Na2CO3 gives how many grams of CO2?
=9.85×44/105.98
=4.09 g of CO2
No.of moles of CO2=4.09/44
=0.092 moles
1 mole = 22.4 litres
∴0.092 mole of CO2=22.4×0.092
=2.06 litres
Therefore, answer is 2.06 litres
Hope it helps...
1 mole of sodium carbonate give 1 mole of carbon dioxide.
105.98 g of sodium carbonate gives 44 g of carbon dioxide.
Then, 9.85 g of Na2CO3 gives how many grams of CO2?
=9.85×44/105.98
=4.09 g of CO2
No.of moles of CO2=4.09/44
=0.092 moles
1 mole = 22.4 litres
∴0.092 mole of CO2=22.4×0.092
=2.06 litres
Therefore, answer is 2.06 litres
Hope it helps...
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