Question

If 6.3 of NaHCO, are added to 15.0 g of CH.COOH solution, the residue is found to weigh 18.0 g. What is the
mass of CO, released in the reaction ?​

Answer 1

Here, the question is wrong, the right question is given below as follows as:

$\huge{\underline{\red{Question:-}}}$

If 6.3 of $NaHCO_3(s)$, are added to 15.0 g of $CH_3COOH(aq)$ solution, the residue is found to weigh 18.0 g. What is the mass of $CO_2(g)$, released in the reaction ?

$\huge{\underline{\green{Answer:-}}}$

The mass of $CO_2(g)$ released in the reaction is 3.3 grams.

$\huge{\underline{\green{Step-by-Step Explanation:-}}}$

The chemical reaction will be as follow as:

$NaHCO_3(s) + CH_3COOH \longrightarrow CH_3COONa(aq) + H_2O(l) + CO_2(g)$

Molar Masses of compounds:-

$NaHCO_3(s)$ = 84 grams

$CH_3COOH(aq)$ = 60 grams

$CH_3COONa(aq)$ = 82 grams

84 g $NaHCO_3(s)$ + 60 g $CH_3COOH(aq)$ → 82 g $CH_3COONa(aq)$ + 44 g $CO_2(g)$

Moles of $NaHCO_3(s)$ = $\frac{6.3}{84}$ = 0.075 mol

Moles of $CH_3COOH(aq)$ =$\frac{15}{60}$ = 0.25 mol

$\therefore{NaHCO_3}$ is a limited reagent.

Moles of $CO_2(g)$ formed = 0.075 mol

Weight of $CO_2(g)$ = 0.075 × 44 = 3.3 grams