4. If 5.0 mol of both hydrochloric acid and sodium sulfide are mixed and reacted according to the equation below, how many moles of hydrogen sulfide (H2S) are produced?

Respuesta :

Answer:

2.5 moles of [tex]H_{2}S[/tex] are produced.

Explanation:

Balanced equation: [tex]Na_{2}S+2HCl\rightarrow H_{2}S+2NaCl[/tex]

According to balanced equation-

1 mol of [tex]Na_{2}S[/tex] produces 1 mol of [tex]H_{2}S[/tex]

So, 5.0 moles of [tex]Na_{2}S[/tex] produce 5.0 moles of [tex]H_{2}S[/tex]

2 moles of HCl produce 1 mol of [tex]H_{2}S[/tex]

So, 5.0 moles of HCl produce 2.5 moles of [tex]H_{2}S[/tex]

As least number of moles of [tex]H_{2}S[/tex] are produced from HCl therefore HCl is the limiting reagent.

Hence 2.5 moles of [tex]H_{2}S[/tex] are produced.

Moles of hydrogen sulfide (H₂S) are produced are 2.5 moles.

Given:

Moles of hydrochloric acid= 5.0 mol

Balanced equation:

[tex]2HCl+ Na_2S----> H_2S+ 2NaCl[/tex]

From the given equation,

  • 1 mol of Na₂S  produces 1 mol of H₂S

So, 5.0 moles of Na₂S produces 5.0 moles of H₂S

  • 2 moles of HCl produce 1 mol of H₂S

So, 5.0 moles of HCl produce 2.5 moles of H₂S

HCl is the limiting reagent since, least number of moles of H₂S are produced.

Thus,  2.5 moles of H₂S  are produced.

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