Respuesta :
Answer:
The theoretical yield of the hydrogen sulfide is 68.0 grams.
Explanation:
[tex]Al_2S_3(s)+6H_2O(l)\rightarrow 2Al(OH)_3(s)+3H_2S(g)[/tex]
Moles of aluminum sulfide = 4.0 mol
Moles of water = 4.0 mol
According to reaction, 6 moles of water reacts with 1 mole of aluminum sulfide,then 4 moles of water will react with :
[tex]\frac{1}{6}\times 4 mol=1.5 mol[/tex] of aluminum sulfide
1.5 moles aluminum sulfide < 4 moles aluminum sulfide
This means that water is present in limiting amount and aluminum sulfide is in excess amount.So, amount of hydrogen sulfide will depend upon moles of water.
According to reaction, 6 moles of water gives with 3 mole of hydrogen sulfide,then 4 moles of water will give :
[tex]\frac{3}{6}\times 4 mol=2.0 mol[/tex] of hydrogen sulfide
Mass of hydrogen sulfide:
2.0 mol × 34 g/mol = 68.0 g
The theoretical yield of the hydrogen sulfide is 68.0 grams.
The theoretical yield of H₂S (in moles) obtained from the reaction is 2 moles
We'll begin by determining the limiting reactant.
Al₂S₃ + 6H₂O —> 2Al(OH)₃ + 3H₂S
From the balanced equation above,
1 mole of Al₂S₃ reacted with 6 moles of H₂O.
Therefore,
4 moles of Al₂S₃ will react with = 4 × 6 = 24 moles of H₂O
From the above calculation, we can see that a higher amount of H₂O (i.e 24 moles) than what was given (i.e 4 moles) is needed to react completely with 4 moles of Al₂S₃.
Therefore, H₂O is the limiting reactant while Al₂S₃ is the excess reactant.
Finally, we shall determine the theoretical yield of H₂S. This can be obtained as follow:
From the balanced equation above,
6 moles of H₂O reacted to produce 3 moles of H₂S.
Therefore,
4 moles of H₂O will react to produce = (4 × 3) / 6 = 2 moles of H₂S.
Thus, the theoretical yield of H₂S is 2 moles
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