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What volume of 0.750 M hydrochoric acid solution can be prepared from the HCl produced by the reaction of25.0 $\mathrm{g}$ of NaCl with excess sulfuric acid?$\mathrm{NaCl}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(l) \longrightarrow \mathrm{HCl}(g)+\mathrm{NaHSO}_{4}(s)$

0.570 L HCl solution

00:31

Sisi G.

Chemistry 101

Chapter 4

Stoichiometry of Chemical Reactions

Chemical reactions and Stoichiometry

Rice University

Drexel University

University of Kentucky

Lectures

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for this next question, the chemical reaction that we are considering is sodium chloride reacting with sulfuric acid producing hydrochloric acid and sodium hydrogen sulphate or sodium. By sulfate, it is balanced as is with one sodium, one sodium, one chlorine, one chlorine to hydrogen to hydrogen and then a sulfate. And this question wants to know the volume of 0.750 molar hydrochloric acid that can be prepared from 25.0 g of sodium chloride. So let's do what we normally do. And because our target compound is hydrochloric acid, let's calculate the moles hydrochloric acid that can be produced. If we have 25 g sodium chloride, we can divide it by the molar mass sodium chloride to get mold sodium chloride. Then we use the strike geometric relationship, 1 to 1 to convert the mold sodium chloride into molds hydrochloric acid and we get 10.4 to 78 moles hydrochloric acid because more clarity is defined as the molds hydrochloric acid divided by the volume of the solution. We can rearrange this equation to calculate the volume of the solution by simply taking the molds hydrochloric acid that we just determined divided by the target polarity, which is the 0.750 Mueller. This gives us 0.570 leaders of the 0.750 molar hydrochloric acid solution that could be produced.

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