When $8.4g$ of $NaHC{O}_3$ is added to a solution of $C{H}_{3}COOH$ weight $20g$.It is observed that $4.4g$ of $C{O}_2$ is released into atmosphere and a residue is left behind.Calculate the mass of residue by applying law of conservation of mass

The mass of residue, is ${CH}_{3}COONa$ solution, should be $24$g of ${CH}_{3}COONa$ solution.

In this reaction, sodium bicarbonate reacts with acetic acid to produce sodium acetate, carbon dioxide and water

${NaHCO}_3+{CH}_{3}COOH⟶{CH}_{3}COONa+{CO}_2+H_{2}O$

According to law of conversation of masses,

total masses of reactants $=$ total masses of products

reactants are $8.4$g ${NaHCO}_3$ and $20$g ${CH}_{3}COOH$.

total mass of reactants is $=28.4$g

mass of residue $=$ ($28.4$g reactants mass) $-$ ($4.4$g ${CO}_2$)

$=$ $24.0$g residue (sodium acetate solution)

mass of residue is thus $24$ grams.