I am trying to understand if my reasoning is right, as my exam is getting closer and I would really appreciate any feedback, I'd be very pleased to know if I have made any mistakes in solving this exercise.
The problem
A solution made of benzene-hydrochloric acid at 5.5% in weight needs to be purified by lowering the amount of acid to 3% in weight. In order to do that the above-mentioned solution is sent to a machinery, whereas it is dipped down in an aqueous solution of sodium hydroxyde. Part of the hydrochloric acid is solubilized in the aqueous current and reacts to the sodium hydroxyde by wearing it out according to the following formula:
$$\begin{aligned} HCl+NaOH \rightarrow NaCl + H_2O\end{aligned}$$
Two liquid currents come out of the machinery: the first one contains the remaining hydrochloric acid and all of benzene, the second one contains water and sodium chloride. If we know that the sodium hydroxyde concentration in water is at 4%, calculate the flow rate of the aqueous current needed to process 137 lb/h of benzene-hydrochloric acid solution.
My solution to the problem
First thing from the benzene-hydrochloridric acid solution I managed to calcuate the incoming and outcoming flow rate of the hydrochloridric acid, respectively 7.54 lb/h and 4.52 lb/h. The difference between these two amounts is actually the amount of hydrochloric acid reacting to the sodium hydroxyde. Due to stoichiometry the amount of hydrochroric acid involved in the reaction is equal to the amount of sodium hydroxyde. Therefore given the data of my problem, the newly found sodium hydroxyde amount is the 4% of the aqueous solution of the sodium hydroxyde going in the machinery. So by proportion I can calculate the water amount of the solution and then the total flow rate of the solution:
$$\begin{aligned} 3.02 : 4\% = x : 96\% \Rightarrow x = 75.5\%\end{aligned}$$.
