Question

A sample contains a mixture of NaHCO$_3$ and Na$_2$CO$_3$ is added to 15.0 g of the sample yielding 11.0 of NaCl. What percent of the sample is Na$_2$CO$_3$?
Reactions are:
(i) Na$_2$CO$_3$ + 2HCl $\rightarrow$ 2NaCl + CO$_2$ + H$_2$O
(ii) NaHCO$_3$ + HCl $\rightarrow$ NaCl + CO$_2$ + H$_2$O
Molecular weight of NaCl, NaHCO$_3$ and Na$_2$CO$_3$ is 58.5, 84, and 106 g/mol respectively.

Answer 1

Hint: 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 Na$_2$CO$_3$, and the weight of NaHCO$_3$ i.e. (15-x) g.
If we see the given reactions, we can say that there are some moles of NaCl produced by Na$_2$CO$_3$, and NaHCO$_3$.
Thus, moles of NaCl produced in the reaction of Na$_2$CO$_3$ = x/106 mol, here 106 represents the molecular weight of Na$_2$CO$_3$.
Thus, moles of NaCl produced in the reaction of NaHCO$_3$ = (15-x)/84 mol, here 84 represents the molecular weight of NaHCO$_3$.
From the reactions, it can be seen that the 2 moles of NaCl produced by Na$_2$CO$_3$.
So, it can be written as $\dfrac{2x}{106}$ + $\dfrac{15-x}{84}$ = 0.188
Therefore, we get x = 13.5 g of Na$_2$CO$_3$.
Thus, 15-x = 15-13.5= 13.6 g of NaHCO$_3$.
Now, we will calculate the percentage of sodium carbonate in the sample.
% of Na$_2$CO$_3$ = $\dfrac{1.35}{15}$ $\times$ 100= 9.0% of Na$_2$CO$_3$.
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.