For the reaction H₂(g)+ 1/2 O₂(g)→ H₂O(g)Δ_rH° = -241.8 kJ/mol, What quantity of heat,in kJ,evolved when a 72.0 g mixture containing equal parts of H₂ and O₂,by mass,is burned? A)8630 kJ B)4860 kJ C)1088 kJ D)272 kJ E)544 kJ

Question

Quizplus · Accepted Answer

First, determine the molar mass of H2 and O2 to find out how many moles of each are in 36.0 g (since the mixture is equal parts by mass, each element accounts for half of the total mass, 72.0 g). The molar mass of H2 is approximately 2.02 g/mol, and for O2, it's approximately 32.00 g/mol. Thus, for H2, $36.0 \, \text{g} \div 2.02 \, \text{g/mol} = 17.82 \, \text{mol}$, and for O2, $36.0 \, \text{g} \div 32.00 \, \text{g/mol} = 1.125 \, \text{mol}$. The reaction requires 2 moles of H2 for every 1 mole of O2, but here O2 is the limiting reagent. Since the reaction consumes O2 at a rate of 0.5 moles of O2 per mole of H2O produced, 1.125 moles of O2 will produce 2.25 moles of H2O. Given the ΔH° of -241.8 kJ/mol for the reaction, the total heat evolved is $2.25 \, \text{mol} \times -241.8 \, \text{kJ/mol} = -544.05 \, \text{kJ}$, which rounds to -544 kJ.

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