Respuesta :
The reaction of sodium sulfide with hydrogen sulfide can be shown as
[tex]H_2S + 2 NaOH = Na_2S + 2 H_2O
\\ Molar mass of H_2S = 34 g /mol
\\ Mole of H_2S = mass /molar mass
\\ = 1.8 /34 = 0.051 mole
\\ Mole of NaOH = 2 * 0.051 = 0.102 mole
\\ Theoritical mole of NaOH = mass /molar mass
\\ = 2.4 /40 = 0.06 mole[/tex]
Sodium hydroxide is completely reacting and thus act as limiting reactant
[tex]Hence, mole of Na_2S required = 0.06 /2 = 0.03 mole
\\ Mass of Na_2S = 0.03 * 78 = 2.31 g[/tex]
Thus, 2.31 g Na2S produced.
Answer: The mass of sodium sulfide formed will be 2.18 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For hydrogen sulfide:
Given mass of hydrogen sulfide = 1.80 g
Molar mass of hydrogen sulfide = 34.1 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of hydrogen sulfide}=\frac{1.80g}{34.1g/mol}=0.053mol[/tex]
- For NaOH:
Given mass of NaOH = 2.40 g
Molar mass of NaOH = 40 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of NaOH}=\frac{2.40g}{40g/mol}=0.06mol[/tex]
The chemical equation for the reaction of hydrogen sulfide and NaOH follows:
[tex]H_2S+2NaOH\rightarrow Na_2S+2H_2O[/tex]
By Stoichiometry of the reaction:
2 moles of NaOH reacts with 1 mole of hydrogen sulfide
So, 0.06 moles of NaOH will react with = [tex]\frac{1}{2}\times 0.06=0.03mol[/tex] of hydrogen sulfide
As, given amount of hydrogen sulfide is more than the required amount. So, it is considered as an excess reagent.
Thus, NaOH is considered as a limiting reagent because it limits the formation of product.
By Stoichiometry of the reaction:
2 moles of NaOH produces 1 mole of sodium sulfide
So, 0.06 moles of NaOH will produce = [tex]\frac{1}{2}\times 0.06=0.03mol[/tex] of sodium sulfide
Now, calculating the mass of sodium sulfide from equation 1, we get:
Molar mass of sodium sulfide = 78.04 g/mol
Moles of sodium sulfide = 0.03 moles
Putting values in equation 1, we get:
[tex]0.03mol=\frac{\text{Mass of sodium sulfide}}{78.04g/mol}\\\\\text{Mass of sodium sulfide}=(0.03mol\times 78.04g/mol)=2.3412g[/tex]
To calculate the experimental yield of sodium sulfide, we use the equation:
[tex]\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100[/tex]
Percentage yield of sodium sulfide = 93.0 %
Theoretical yield of sodium sulfide = 2.3412 g
Putting values in above equation, we get:
[tex]93.0=\frac{\text{Experimental yield of sodium sulfide}}{2.3412g}\times 100\\\\\text{Experimental yield of sodium sulfide}=\frac{93.0\times 2.3412}{100}=2.18g[/tex]
Hence, the mass of sodium sulfide formed will be 2.18 grams.
