Question

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​

Answer 1

Answer:

This is a balanced equation

Explanation:

Answer 2

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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