Question: \[NaCl{\text{ }} + {\text{ }}{H_2}S{O_4} \to {\text{ }}HCl{\text{ }} + {\text{ }}N{a_2}S{O_4}\] If e...

Question

$NaCl{\text{ }} + {\text{ }}{H_2}S{O_4} \to {\text{ }}HCl{\text{ }} + {\text{ }}N{a_2}S{O_4}$ If excess sulfuric acid ( ${H_2}S{O_4}$ ) reacts with $30.0$ grams of sodium chloride, how many grams of hydrogen chloride are produced?

Answer 1

Solution

The response of sodium chloride and sulfuric acid produces hydrochloric fuel line as a through-product. The fueloline hydrogen chloride escapes from the medium so the response can't take the region in a backward direction, and as a consequence the response is irreversible.

Complete answer:
The reaction between sodium chloride and sulfuric acid is a single displacement reaction that leads to the product sodium sulphate and hydrochloric acid. If we've stable sodium chloride and focused sulfuric acid then an acid/base response occurs. Steamy fumes of hydrogen chloride fuel line are observed. The different product is sodium hydrogen sulphate $NaHS{O_4}$ . This paperwork has the idea of the laboratory coaching of hydrogen chloride fueloline.
First, we should see that the equation is balanced. Then, we must convert the given grams to moles of reactant. We use that to decide the moles of the product from the equation. Finally, we convert the one’s product moles again into grams.
This equation isn't always balanced, however, due to the fact an extra acid is given, the most effective pertinent reality is the ratio of sodium chloride to hydrogen chloride, which stays $1:1$ . The balanced response is:
$2NaCl{\text{ }} + {\text{ }}{H_2}S{O_4} \to {\text{ 2}}HCl{\text{ }} + {\text{ }}N{a_2}S{O_4}$
Moles of $NaCl$ :

= \dfrac{{30.0}}{{58.5}} \\\ = 0.513 \\\

$0.513\;moles{\text{ }}of\;HCl = 0.513 \times 36.5 = 18.7g$

Note:
An unmarried displacement response which is likewise referred to as an unmarried substitute response is a sort of oxidation-discount chemical response whilst an ion or detail moves out of a compound, i.e., one detail is changed through the opposite in a compound. Double displacement reactions take region in most cases in aqueous answers in which the ions precipitate and alternate ions take region.