Question

16.8g sodium hydrogen carbonate is added to 12.0g of acetic acid.the mass of carbon dioxide produced in the reaction is

Answer 1

The equation for the reaction of sodium hydrogen carbonate and acetic acid is as follows:

CH₃COOH + NaHCO₃ --------------> CH₃COONa + H₂O + CO₂

Next: Find the limiting reagent between CH₃COOH and Na2HCO₃

Find the moles of CH₃COOH

moles = mass/molar mass
           = 12/60
           = 0.2 moles

Find the moles of NaHCO₃

mass/molar mass = 16.8/84
                             = 0.2 moles

Since the mole ratio of the CH₃COOH and Na2HCO₃ is 1 : 1 in the equation above, and we see both use 0.2 moles for the equation, then there is no limiting reagent. Both compounds are completley utilized:

You can use either of the two reagents in this case to calculate the mass o f CO₂ produced.

The mole ratio between CH₃COOH and the CO₂ produced is 1:1 as well

This means the moles of the CO₂ produced will also be 0.2 moles

Use this mole to calculate the mass of CO₂ produced

Mass = mass x molar mass
       molar mass CO₂ = 44  and moles = 0.2 moles

Therefore mass= 0.2 moles x 44g/mol
                         = 8.8 g

The mass of CO₂ produced thus is 8.8 g

Answer 2

Balanced chemical equation of the reaction is
CH3COOH+NAHCO3->CH3COONA+ H2O+CO2
Which shows 1, 1 mole of CH3COOH AND NAHCO3 WILL GIVE 1 MOLE OF CO2.
16.8g NAHCO3 HAS MOLES=16.8g/84=0.2 moles
12g CH3COOH has 9moles=12g/60=0.2moles
As 1,1 mole of both give 1 mole of CO2 by this 0.2 moles of both will give 0.2 mole of CO2.
Molar mass of CO2=44
So mass of CO2 produced will be
Mass=molar mass×moles
Mass=44×0.2=8.8g