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Hydrochloric acid can be prepared using the reaction described by the chemical equation: 2 NaCl(s) + H2SO4(l) ----> 2 HCl(g) + Na2 SO4(s). How many grams of HCl can be prepared from 393 g of H2SO4 and 4.00 moles of NaCl?

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Answer:

145.84 g

Explanation:

  • 2 NaCl(s) + H₂SO₄(l) → 2 HCl(g) + Na₂SO₄(s)

To answer this problem first we need to determine the limiting reactant, to do so we convert 393 g of H₂SO₄ into moles using its molar mass:

  • 393 g H₂SO₄ ÷ 98 g/mol = 4.01 mol

4.01 moles of H₂SO₄ would react completely with (4.01 * 2) 8.02 moles of NaCl. There are not as many NaCl moles, so NaCl is the limiting reactant.

Now we calculate how many HCl moles are formed, using the number of moles of the limiting reactant and the stoichiometric coefficients of the reaction:

  • 4.00 mol NaCl * [tex]\frac{2molHCl}{2molNaCl}[/tex] = 4.00 mol HCl

Finally we convert HCl moles into grams, using its molar mass:

  • 4.00 mol HCl * 36.46 g/mol = 145.84 g

According to the law of chemical combination, the mass of the element remains the same after the reaction.

The law depends on the following

  • Mass

The correct answer is 145.84g

Before solving the question, we have to find the limiting reagent to the reaction. The compound which decide the production of the reaction is called limiting reagent.

The moles is as follows:-

[tex]\frac{393g H_2SO_4}{98g/mol}\\\\ = 4.01 mol[/tex]

4.01 moles [tex]H_2SO_4[/tex] would react with the 2 moles of NaCl.

Hence, [tex](4.01 * 2 ) = 8.02[/tex]

According to the law the moles of the following.

4.00 mole NaCl *  = 4.00 mole HCl

Convert the moles into grams the solution is as follows:-

[tex]4.00 mol HCl * 36.46 g/mol = 145.84 g[/tex]

For more information, refer to the link:-

https://brainly.com/question/25305623

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