If `6.3 g` of `NaHCO_(3)` are added to `15.0 g` `CH_(3)COOH` solution, the residue is found of weight `18.0 g`. What is the mass of `CO_(2)` released in the reaction?

Options

A. `4.5 g`
B. `3.3 g`
C. `2.6 g`
D. `2.8 g`

Answer

B

Explanation

The chemical reaction leading to products is
`"sodium hydrogen carbonate"+"ethanoic acid" to underset("residue left")("sodium ethanoic solution")+underset("released")("carbon dioxide")`
`"Mass of reactants"=(6.3+15.0)=21.3g`
`"Mass of products"="Mass of residue"+"Mass of carbon dioxide released"`
=18.0+x g
According to law of conservation of mass.
Mass of reactants=Mass of products
21.3g=(18.0+x)g
or x=21.3-18.0=3.3g
Mass of carbon dioxide released=3.3g