Question

6 gram acetic acid is treated with excess of sodiumbicarbonate. Calculate the mass in gram of the gas released in this reaction.​

Answer 1

Answer:

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Explanation:

The mass of residue, is CH

3

COONa solution, should be 24g 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 8.4g NaHCO

3

and 20g CH

3

COOH.

total mass of reactants is =28.4g

mass of residue = (28.4g reactants mass) − (4.4g CO

2

)

= 24.0g residue (sodium acetate solution)

mass of residue is thus 24 grams

Answer 2

Answer:

4.4 g of $CO_{2}$ gas is released when 6 g acetic acid is treated with an excess of sodium bicarbonate.

Explanation:

Given mass of acetic acid = 6 g

Molar mass of acetic acid($CH_{3}COOH$) = (2 × 12) + (4 × 1) + (2 × 16) = 60 g/mol

Number of moles of acetic acid = Given mass ÷ molar mass

                                                     = $\frac{6}{60}$ = 0.1 moles

Acetic acid reacts with sodium bicarbonate according to the following balanced equation:

$NaHCO_{3} + CH_{3}COOH$ → $CH_{3} COONa + CO_{2} +H_{2} O$

The gas released is carbon dioxide gas

1 mole of acetic acid reacts to give 1 mole of carbon dioxide.

⇒0.1 moles of acetic acid reacts to give 0.1 moles of carbon dioxide.

Molar mass of $CO_{2}$ = (1 × 12) + (2 × 16) = 44 g/mol

⇒Grams of $CO_{2}$ = number of moles × molar mass

                          = 0.1 × 44 = 4.4 g

Therefore, 4.4 g of $CO_{2}$ gas is released when 6 g acetic acid is treated with an excess of sodium bicarbonate.