Question

One mole of a mixture of CO and \[C{{O}_{2}}\] ​ requires exactly 20 g of NaOH in solution for complete conversion of all the \[C{{O}_{2}}\]into \[N{{a}_{2}}C{{O}_{3}}\] ​. If the mixture (one mole) is completely oxidised to \[C{{O}_{2}}\]​, then weight of NaOH to be used is:
a. 60 g
b. 80 g
c. 40 g
d. 20 g

Answer 1

Hint: To be able to solve this question effectively and quickly, you need to remember the reaction between Sodium Hydroxide and Carbon Dioxide that results in the production of Sodium Carbonate. Balance all stoichiometric coefficients of this reaction and then try to calculate the number of moles of \[C{{O}_{2}}\] initially present, which will also then help you calculate the number of moles of CO in the mixture.

Complete step by step answer:
To solve this question, we need to first understand the law of constant combination which heavily influences the solution.
In chemistry, the law of definite proportion, sometimes called Proust's law, or law of constant composition states that a given chemical compound always contains its component elements in fixed ratio and does not depend on its source and method of preparation.
With this in mind, let us now look at the reaction between Sodium Hydroxide and Carbon Dioxide that results in the production of Sodium Carbonate. We find that the reaction in question is,
\[2NaOH+C{{O}_{2}}\to N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O\]
Now, we know that the molar mass of NaOH is 40.
Therefore, we observe 80 grams of NaOH is required for the complete conversion of 1 mole of Carbon Dioxide into Sodium Carbonate.
Since, there are only 20 grams of NaOH present in the given solution.
Therefore, we can conclude that only 0.25 moles of \[C{{O}_{2}}\] is present in the given mixture of \[C{{O}_{2}}\] and CO.
Now, if the other 0.75 moles in the mixture (originally of CO) are oxidised to CO, then the total number of moles of \[C{{O}_{2}}\] is now 1.
Then, as we saw in the above reaction, 80 grams of NaOH are required for its complete conversion to \[N{{a}_{2}}C{{O}_{3}}\].
Thus, we can safely conclude that the answer to this question is b) 80 grams.

Note: Although very useful in the foundation of modern chemistry, the law of definite proportions is not universally true. There exist non-stoichiometric compounds whose elemental composition can vary from sample to sample. Such compounds follow the law of multiple proportions.