
A $\,1.50g\,$ sample of potassium bicarbonate having $\,80\% \,$ purity is strongly heated. Assuming the impurity to be thermally stable, the loss in weight of the sample, on heating, is ____________.
A.$\,3.72g\,$
B.$\,0.72g\,$
C.$\,0.372g\,$
D.$\,0.186g\,$
Answer
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Hint: If heated beyond a certain temperature, potassium bicarbonate decomposes to potassium carbonate. It yields water and carbon dioxide as by-products. Potassium bicarbonate is a monopotassium salt of carbonic acid.
Complete step by step answer:
Let us first look into the chemical reaction when potassium bicarbonate is heated;
$\,2KHC{O_3}\xrightarrow{{Heat}}{K_2}C{O_3} + {H_2}O + C{O_2}\,$
The mass of above compounds are as follows;
$\,2KHC{O_3} = 200g\,$,$\,{K_2}C{O_3} = 138g\,$,$\,{H_2}O = 18\,$,$\,C{O_2} = 44\,$
While heating strongly water and carbon dioxide will escape but potassium carbonate will not escape as it is thermodynamically stable.
Now, while analysing the question we understand that;
$\,80\% \,$ of $\,1.50g\,$ means $\,80 \times \dfrac{{1.50}}{{100}}\, = 1.2g\,$.
Which means, out of the given mass of potassium bicarbonate only $\,1.2g\,$is the pure content of potassium bicarbonate and the remaining is impurity. So, from the above equation we can see that;
When $\,200g\,$ of $\,KHC{O_3}\,$ is heated, $\,62g(44 + 18)\,$ is the loss in weight of the sample as \[\,44g\,\] and $\,18g\,$ are the respective molecular masses of $\,C{O_2}\,$ and $\,{H_2}O\,$ which are the impurities here.
This was the mass of impurities for two molecules of $\,KHC{O_3}\,$ heated. So, for one molecule ($\,(100g)\,$ this becomes;
$\,x = \dfrac{{62}}{{200}} \times 100\,$
$\, = 31g\,$
So, when $\,1.2g\,$( which is the amount of pure substance present here) is heated, the loss in weight is;
$ \Rightarrow x = \dfrac{{31}}{{100}} \times 1.2\,$or $\,\dfrac{{62}}{{200}} \times 1.2\,$
$\, = 0.372g\,$
Hence, the answer to the above question is option C.
Additional information: Similar to sodium bicarbonate $\,(\,$baking soda$\,)\,$, potassium bicarbonate is alkaline in nature. As a result, in baking, this compound is a source of carbon dioxide for leavening. It will supplement those with a reduced sodium diet with baking soda sodium bicarbonate, and it is an ingredient in low sodium baking powder. It is widely used as an antacid since it has the ability to neutralise gastric acid. Potassium bicarbonate is used in some dry chemical fire extinguishers as a fire control agent. Also, it is commonly used in crops, especially for acidic soil neutralisation.
Note:
Decomposition of the bicarbonate occurs between $\,100\,$ and $\,120\,$ degree Celsius. This chemical decomposition is used to prepare high quality potassium carbonate in industries.
Complete step by step answer:
Let us first look into the chemical reaction when potassium bicarbonate is heated;
$\,2KHC{O_3}\xrightarrow{{Heat}}{K_2}C{O_3} + {H_2}O + C{O_2}\,$
The mass of above compounds are as follows;
$\,2KHC{O_3} = 200g\,$,$\,{K_2}C{O_3} = 138g\,$,$\,{H_2}O = 18\,$,$\,C{O_2} = 44\,$
While heating strongly water and carbon dioxide will escape but potassium carbonate will not escape as it is thermodynamically stable.
Now, while analysing the question we understand that;
$\,80\% \,$ of $\,1.50g\,$ means $\,80 \times \dfrac{{1.50}}{{100}}\, = 1.2g\,$.
Which means, out of the given mass of potassium bicarbonate only $\,1.2g\,$is the pure content of potassium bicarbonate and the remaining is impurity. So, from the above equation we can see that;
When $\,200g\,$ of $\,KHC{O_3}\,$ is heated, $\,62g(44 + 18)\,$ is the loss in weight of the sample as \[\,44g\,\] and $\,18g\,$ are the respective molecular masses of $\,C{O_2}\,$ and $\,{H_2}O\,$ which are the impurities here.
This was the mass of impurities for two molecules of $\,KHC{O_3}\,$ heated. So, for one molecule ($\,(100g)\,$ this becomes;
$\,x = \dfrac{{62}}{{200}} \times 100\,$
$\, = 31g\,$
So, when $\,1.2g\,$( which is the amount of pure substance present here) is heated, the loss in weight is;
$ \Rightarrow x = \dfrac{{31}}{{100}} \times 1.2\,$or $\,\dfrac{{62}}{{200}} \times 1.2\,$
$\, = 0.372g\,$
Hence, the answer to the above question is option C.
Additional information: Similar to sodium bicarbonate $\,(\,$baking soda$\,)\,$, potassium bicarbonate is alkaline in nature. As a result, in baking, this compound is a source of carbon dioxide for leavening. It will supplement those with a reduced sodium diet with baking soda sodium bicarbonate, and it is an ingredient in low sodium baking powder. It is widely used as an antacid since it has the ability to neutralise gastric acid. Potassium bicarbonate is used in some dry chemical fire extinguishers as a fire control agent. Also, it is commonly used in crops, especially for acidic soil neutralisation.
Note:
Decomposition of the bicarbonate occurs between $\,100\,$ and $\,120\,$ degree Celsius. This chemical decomposition is used to prepare high quality potassium carbonate in industries.
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