BRAINLIEST PLEASEEE HELPLP 5. In a lab experiment, 2.5 grams of sodium bicarbonate is heated and decomposed into sodium carbonate, carbon dioxide, and water vapor when heated. The actual yield of
sodium carbonate produced in the experiment is 2.04 grams. The theoretical yield of
each product is recorded in the data table below.
Using this data, determine the percent yield for sodium carbonate?
(Round Your Answer to the Nearest Whole Number)

Explanation:

Sodium bicarbonate,

NaHCO

3

, will decompose to form sodium carbonate,

Na

2

CO

3

, water, and carbon dioxide,

CO

2

2

NaHCO

3(s]

→

Na

2

CO

3(s]

+

CO

2(g]

+

H

2

O

(g]

Notice that you have a

2

:

1

mole ratio between sodium bicarbonate and sodium carbonate. This means that the reaction will produce half as many moles of the latter than whatever number of moles of the former underwent decomposition.

Use sodium carbonate's molar amss to determine how many moles you'd get in that sample

0.685

g

⋅

1 mole NaHCO

3

84.007

g

=

0.008154 moles NaHCO

3

Now, if the reaction were to have a

100

%

yield, it would produce

0.008154

moles NaHCO

3

⋅

1 mole Na

2

CO

3

2

moles NaHCO

3

=

0.004077 moles Na

2

CO

3

Use the molar mass of sodium carbonate to determine how many grams would contain this many moles

0.004077

moles

⋅

105.99 g

1

mole

=

0.4321 g Na

2

CO

3Sodium bicarbonate,

NaHCO

3

, will decompose to form sodium carbonate,

Na

2

CO

3

, water, and carbon dioxide,

CO

2

2

NaHCO

3(s]

→

Na

2

CO

3(s]

+

CO

2(g]

+

H

2

O

(g]

Notice that you have a

2

:

1

mole ratio between sodium bicarbonate and sodium carbonate. This means that the reaction will produce half as many moles of the latter than whatever number of moles of the former underwent decomposition.

Use sodium carbonate's molar amss to determine how many moles you'd get in that sample

0.685

g

⋅

1 mole NaHCO

3

84.007

g

=

0.008154 moles NaHCO

3

Now, if the reaction were to have a

100

%

yield, it would produce

0.008154

moles NaHCO

3

⋅

1 mole Na

2

CO

3

2

moles NaHCO

3

=

0.004077 moles Na

2

CO

3

Use the molar mass of sodium carbonate to determine how many grams would contain this many moles

0.004077

moles

⋅

105.99 g

1

mole

=

0.4321 g Na

2

CO

3