How much energy is produced when 93.5 grams of oxygen react with 13.2 grams of hydrogen in the following reaction:

Question

2H2+o2-->2 h2o triangleH=-572kJ

Answer 1

To answer this question, we need to first determine which reactant is limiting.

1. Calculate the moles of each reactant:
- Moles of O2 = 93.5 g / 32.00 g/mol = 2.92 mol
- Moles of H2 = 13.2 g / 2.016 g/mol = 6.56 mol

2. Use the mole ratio from the balanced chemical equation to determine which reactant is limiting:
- According to the equation, 2 moles of H2 react with 1 mole of O2
- Therefore, if we assume all of the O2 reacts, we would need 2 * 2.92 = 5.84 moles of H2
- Since we only have 6.56 moles of H2, it is in excess and O2 is limiting.

3. Calculate the amount of energy produced:
- Since we know O2 is limiting, we can use the stoichiometry from the equation to find the moles of water produced:
2 mol of H2O are produced per 1 mol of O2 reacted, so
- Moles of H2O produced = 1/2 * 2.92 mol = 1.46 mol

- Finally, we can use the standard enthalpy change (triangleH) given in the question to calculate the energy produced:
- Energy produced = triangleH * moles of H2O produced
- Energy produced = -572 kJ/mol * 1.46 mol = -835.92 kJ

Therefore, the energy produced by the reaction is -835.92 kJ (the negative sign indicates that the reaction is exothermic, or releases energy).