Question

16.8g of sodium hydrogen carbonate is added to 12.0g of ethanoic acid. The residue left weighed 20.0g. The amount of carbon dioxide formed is

Answer 1

Answer:

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

Explanation: