How many liters of O2 are needed to react completely with 45.0 L of H2S at STP? (2 points) 2H2S (g) + 3O2 (g) 2SO2 (g) + 2H2O (g)
The balanced chemical reaction is given as:
2H2S (g) + 3O2 (g) 2SO2 (g) + 2H2O (g)
We are given the volume at STP of H2S to be used for the reaction. This would be the starting point for the calculations. We do as follows:
45.0 L H2S ( 1 mol / 22.4 L) ( 3 mol O2 / 2 mol H2S ) ( 22.4 L / 1 mol ) = 67.5 L O2 needed
The volume of O₂ needed for the reaction is 67.5 L
From the balanced equation above,
2 L of H₂S required 3 L of O₂
2 L of H₂S required 3 L of O₂.
Therefore,
45 L of H₂S will require = (45 × 3) / 2 = 67.5 L of O₂
Thus, 67.5 L of O₂ is needed for the reaction
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