The volume of CO₂ gas obtained at STP on heating 8.4 g NaHCO3 is (molar mass of NaHCO3 = 84 g/mol)
2NaHCO, Na₂CO₂ + H₂O + CO₂
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The volume of CO₂ gas obtained at STP on heating 8.4 g NaHCO3 is (molar mass of NaHCO3 = 84 g/mol)
2NaHCO, Na₂CO₂ + H₂O + CO₂
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Answer:
This is a balanced equation
Explanation:
Verified answer
Answer:-
2NaHCO3(8.4g)Na2CO+H2O+CCo2↑
Here,
2moleofNaHCO3gives1moleofCO2gas
so,(84/8.4)moles of NaHCO3 gives=1/2×(84/8.4)moles
Since,1moles(gives)=22.4L
∴ 1/2×(84/8.4)g
= 1/2×(84/8.4)×22.4L
=1.12L
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