For the reaction H₂(g)+ 1/2 O₂(g)→ H₂O(g)Δ_rH° = -241.8 kJ/mol, What mass of H₂(g)is required to liberate 1.00 × 10³ kJ of heat? A)66.2 g B)4.17 g C)8.34 g D)2.05 g E)16.7 g

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Quizplus · Accepted Answer

Given that ΔH° = -241.8 kJ/mol for the reaction of H₂(g) with 1/2 O₂(g) to form H₂O(g), to liberate 1.00 × 10³ kJ (or 1000 kJ) of heat, we first calculate the moles of H₂ needed using the ratio from the reaction's ΔH° value. $$1000 \, \text{kJ} \times \frac{1 \, \text{mol H₂}}{241.8 \, \text{kJ}} = 4.134 \, \text{mol H₂}$$The molar mass of H₂ (H = 1.01 g/mol) is approximately 2.02 g/mol, so:$$4.134 \, \text{mol} \times 2.02 \, \text{g/mol} = 8.35 \, \text{g}$$Therefore, approximately 8.34 g of H₂(g) is required to liberate 1000 kJ of heat.

For the Reaction H2(g)+ 1/2 O2(g)→ H2O(g)ΔrH° = -241

For the Reaction H₂(g)+ 1/2 O₂(g)→ H₂O(g)Δ_rH° = -241