Question

If 6.3 g of $NaHCO_{3}$ are added to 15.0 g $CH_{3}COOH$ solution, the residue is found to weigh 18.0 g. What is the mass of $CO_{2}$ released in the reaction?

Answer 1

"Molecular weights:

$NaH{ CO }_{ 3 } = 84$

${ CH }_{ 3 }COOH = 60$

${ CH }_{ 3 }COONa = 82$

${ CO }_{ 2 } = 44$

"84 grams" $NaH{ CO }_{ 3 }$reacts with "60 grams" of ${ CH }_{ 3 }COOH$to form "82 grams" of ${ CH }_{ 3 }COONa$and "44 grams" of ${ CO }_{ 2 }$.

Given, 6.3 grams of $NaH{ CO }_{ 3 }$reacts with 15 grams of {$CH }_{ 3 }COOH$.

"Number of moles" of $NaH{ CO }_{ 3 } =\quad \frac { Given\quad Weight }{ Molecular\quad Weight }$

$=\quad \frac { 6.3 }{ 84 }$

= 0.075 moles

"Number of moles" of ${ CH }_{ 3 }COOH =\quad \frac { 15 }{ 60 }$

= 0.25

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

Thus, "0.075 moles" of $NaH{ CO }_{ 3 }$ will reacts with "0.075 moles" of to produce "0.075 moles" of ${ CO }_{ 2 }.$

Weight of 1 mole of ${ CO }_{ 2 }$ = 44 grams

Weight 0.075 moles of ${ CO }_{ 2 } =\quad \left( 0.075\quad \times \quad 0.44 \right)$

=3.3 grams

So, 3.3 grams of ${ CO }_{ 2 }$will be produced in the reaction."

Answer 2

Answer:

The chemical reaction will be:

NaHCO

3

+CH

3

COOH→CH

3

COONa+H

2

O+CO

2

molarmass:

NaHCO

3

=84

CH

3

COOH=60

CH

3

COONa=82

CO

2

=44

84gNaHCO

3

+60gCH

3

COOH→82gCH

3

COONa+44gCO

2

Moles of NaHCO

3

=

84

6.3

=0.075

Moles of CH

3

COOH=

60

15

=0.25

∴NaHCO

3

is the limited reagent.

Moles of CO

2

formed=0.075

weight of CO

2

=0.075×44=3.3g