The following thermochemical equation is for the reaction of hydrogen sulfide(g) with water(l) to form hydrogen(g) and sulfur dioxide(g).
H2S(g) + 2H2O(g) -> 3H2(g) + SO2(g) ΔH = 207 kJ
When 10.4 grams of hydrogen sulfide(g) react with excess water(l),______kJ of energy are _________ (evolved or absorbed)
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To determine the amount of energy evolved or absorbed in the reaction, we need to calculate the energy change for the given amount of reactant.
First, we need to convert the mass of hydrogen sulfide (H2S) to moles. The molar mass of H2S is 34.08 g/mol.
Moles of H2S = mass of H2S / molar mass of H2S
= 10.4 g / 34.08 g/mol
≈ 0.305 mol
From the balanced equation, we can see that 1 mole of H2S produces 3 moles of H2. Therefore, 0.305 moles of H2S will produce:
Moles of H2 = 0.305 mol H2S * 3 mol H2 / 1 mol H2S
= 0.915 mol
Now we can use the given ΔH value of 207 kJ to calculate the energy change for the reaction.
Energy change = ΔH * moles of H2
= 207 kJ * 0.915 mol
≈ 189.105 kJ
Therefore, approximately 189.105 kJ of energy are evolved in the reaction.
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