Question

When 4.2 g of $NaHCO_{3}$ ( sodium hydrogen carbonate) is added to a solution of $CH_{3}COOH$ (acetic acid) weighing 10.0 g, it is observed that 2.2 g $CO_{2}$ is released into atmosphere. The residue is found to weight 12.0 g. Show that three observations are in agreement with the law of conservation of mass.

Answer 1

NaHCO3 + CH3COOH → CH3COONa + H2O + CO2

Molecular weights

NaHCO3 = 84

CH3COOH = 60

CH3COONa = 82

CO2 = 44

84 grams NaHCO3 reacts with 60 grams of CH3COOH to form 82 grams of CH3COONa and 44 grams of CO2.

Given 6.3 grams of NaHCO3 reacts with 15 grams of CH3COOH.

Number of moles of NaHCO3 = given weight/ moleculear weight

= 6.3/84

= 0.075moles

Number of moles of CH3COOH = 15/60

= 0.25

It is to be remember that 1 mole of reactants combines to produce 1 mole of products

Thus 0.075 moles of NaHCO3 will reacts with 0.075 moles of to produce 0.075 moles of CO2.

Weight of 1 mole of CO2 is = 44 grams

Weight 0.075 moles of CO2 = (0.075*44)

= 3.3 grams

So 3.3 grams of CO2 will be produced in the reaction.

Answer 2

            When 4.2 g of $NaHCO_{3}$ ( sodium hydrogen carbonate) is added to a solution of $CH_{3}COOH$ (acetic acid) weighing 10.0 g, it is observed that 2.2 g $CO_{2}$ is released into atmosphere.

            The residue is found to weight 12.0 g. Show that three observations are in agreement with the law of conservation of mass.