
19 grams of a mixture containing $NaHC{O_3}$ and $N{a_2}C{O_3}$ on complete heating liberated 1.12 litres of $C{O_2}$ at STP. The weight of the remaining solid was 15.9g. What is the weight (in g) of $N{a_2}C{O_3}$ in the mixture before heating?
A. 8.4
B. 15.9
C. 4.0
D. 10.6
Answer
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Hint: When a mixture contains $NaHC{O_3}$ and $N{a_2}C{O_3}$, then on heating only $NaHC{O_3}$ produces the carbon dioxide not $N{a_2}C{O_3}$.
So write the decomposition reaction of $NaHC{O_3}$ and find the weight of $NaHC{O_3}$ required to produce 1.12 litres of $C{O_2}$ at STP and subtract this weight from 19 grams to find the weight of $N{a_2}C{O_3}$ in the mixture before heating.
Step by step answer:
We are given that 19 grams of a mixture containing $NaHC{O_3}$ and $N{a_2}C{O_3}$ on complete heating liberated 1.12 litres of $C{O_2}$ at STP. The weight of the remaining solid was 15.9g.
We have to calculate the weight of $N{a_2}C{O_3}$ in the mixture before heating in grams.
When Sodium bicarbonate ($NaHC{O_3}$) is heated, it undergoes thermal decomposition producing sodium carbonate, water and carbon dioxide .
$2NaHC{O_3}\xrightarrow{\Delta }N{a_2}C{O_3} + {H_2}O + C{O_2}$
2 moles of Sodium bicarbonate is required to produce 1 mole of sodium carbonate, 1 mole of water and 1 mole of carbon dioxide. 2 moles of sodium bicarbonate produces 22.4 litres of $C{O_2}$ at STP.
Weight of Sodium (Na) atom= 23g
Weight of Hydrogen atom= 1g
Weight of Carbon atom= 12g
Weight of Oxygen atom= 16g
Weight of one mole of sodium bicarbonate $NaHC{O_3}$ is $\left( {1 \times 23} \right) + \left( {1 \times 1} \right) + \left( {1 \times 12} \right) + \left( {3 \times 16} \right) = 84g$
Weight of 2 moles of $NaHC{O_3}$ is 2×84=168g
168 grams of $NaHC{O_3}$ produces 22.4 litres of $C{O_2}$ at STP
X grams of $NaHC{O_3}$ produces 1.12 litres of $C{O_2}$ at STP
Find the value of x.
$
x = \dfrac{{168 \times 1.12}}{{22.4}} \\
\therefore x = 8.4g \\
$
The weight of $NaHC{O_3}$ required to produce 1,12 litres of $C{O_2}$ is 8.4 grams.
Then the weight of $N{a_2}C{O_3}$ before heating the mixture is $19 - 8.4 = 10.6grams$
The correct option is Option D, the weight (in g) of $N{a_2}C{O_3}$ in the mixture before heating is 10.6.
Note: Sodium bicarbonate is also known as baking soda whereas sodium carbonate is popularly known as washing soda or soda ash. Sodium bicarbonate is made up of sodium, acid and hydrogen whereas sodium carbonate is made up of sodium and acid. Although they sound similar with their names they are different.
So write the decomposition reaction of $NaHC{O_3}$ and find the weight of $NaHC{O_3}$ required to produce 1.12 litres of $C{O_2}$ at STP and subtract this weight from 19 grams to find the weight of $N{a_2}C{O_3}$ in the mixture before heating.
Step by step answer:
We are given that 19 grams of a mixture containing $NaHC{O_3}$ and $N{a_2}C{O_3}$ on complete heating liberated 1.12 litres of $C{O_2}$ at STP. The weight of the remaining solid was 15.9g.
We have to calculate the weight of $N{a_2}C{O_3}$ in the mixture before heating in grams.
When Sodium bicarbonate ($NaHC{O_3}$) is heated, it undergoes thermal decomposition producing sodium carbonate, water and carbon dioxide .
$2NaHC{O_3}\xrightarrow{\Delta }N{a_2}C{O_3} + {H_2}O + C{O_2}$
2 moles of Sodium bicarbonate is required to produce 1 mole of sodium carbonate, 1 mole of water and 1 mole of carbon dioxide. 2 moles of sodium bicarbonate produces 22.4 litres of $C{O_2}$ at STP.
Weight of Sodium (Na) atom= 23g
Weight of Hydrogen atom= 1g
Weight of Carbon atom= 12g
Weight of Oxygen atom= 16g
Weight of one mole of sodium bicarbonate $NaHC{O_3}$ is $\left( {1 \times 23} \right) + \left( {1 \times 1} \right) + \left( {1 \times 12} \right) + \left( {3 \times 16} \right) = 84g$
Weight of 2 moles of $NaHC{O_3}$ is 2×84=168g
168 grams of $NaHC{O_3}$ produces 22.4 litres of $C{O_2}$ at STP
X grams of $NaHC{O_3}$ produces 1.12 litres of $C{O_2}$ at STP
Find the value of x.
$
x = \dfrac{{168 \times 1.12}}{{22.4}} \\
\therefore x = 8.4g \\
$
The weight of $NaHC{O_3}$ required to produce 1,12 litres of $C{O_2}$ is 8.4 grams.
Then the weight of $N{a_2}C{O_3}$ before heating the mixture is $19 - 8.4 = 10.6grams$
The correct option is Option D, the weight (in g) of $N{a_2}C{O_3}$ in the mixture before heating is 10.6.
Note: Sodium bicarbonate is also known as baking soda whereas sodium carbonate is popularly known as washing soda or soda ash. Sodium bicarbonate is made up of sodium, acid and hydrogen whereas sodium carbonate is made up of sodium and acid. Although they sound similar with their names they are different.
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