Question
Question: A sample contains a mixture of NaHCO\(_3\) and Na\(_2\)CO\(_3\) is added to 15.0 g of the sample yie...
A sample contains a mixture of NaHCO3 and Na2CO3 is added to 15.0 g of the sample yielding 11.0 of NaCl. What percent of the sample is Na2CO3?
Reactions are:
(i) Na2CO3 + 2HCl → 2NaCl + CO2 + H2O
(ii) NaHCO3 + HCl → NaCl + CO2 + H2O
Molecular weight of NaCl, NaHCO3 and Na2CO3 is 58.5, 84, and 106 g/mol respectively.
Solution
It is a numerical based question. Consider the data given in the question, and calculate the moles of both the compounds. Then, the percentage of the sample can be calculated by going step by step.
Complete step by step solution:
Now, first we will calculate the moles of NaCl, as we are given with the molecular weight of NaCl i.e. 58.5 g, and the given mass is 11.0 g.
Thus, moles of NaCl = 11.0/58.5 = 0.188 mol
Now, according to the question, consider the x g to be the weight of Na2CO3, and the weight of NaHCO3 i.e. (15-x) g.
If we see the given reactions, we can say that there are some moles of NaCl produced by Na2CO3, and NaHCO3.
Thus, moles of NaCl produced in the reaction of Na2CO3 = x/106 mol, here 106 represents the molecular weight of Na2CO3.
Thus, moles of NaCl produced in the reaction of NaHCO3 = (15-x)/84 mol, here 84 represents the molecular weight of NaHCO3.
From the reactions, it can be seen that the 2 moles of NaCl produced by Na2CO3.
So, it can be written as 1062x + 8415−x = 0.188
Therefore, we get x = 13.5 g of Na2CO3.
Thus, 15-x = 15-13.5= 13.6 g of NaHCO3.
Now, we will calculate the percentage of sodium carbonate in the sample.
% of Na2CO3 = 151.35 × 100= 9.0% of Na2CO3.
In the last we can conclude that the percentage of sodium carbonate in the sample is 9.0%.
Note: Don’t get confused while calculating the percentage of samples. Just solve step by step. The molecular weight, and the given weight are two different terms. Molecular weight is the weight of elements combined in a molecule, whereas given weight is the weight of the molecule used to perform a reaction.