If 10.4 mols of hydrogen, H2, and 9.1 mols of oxygen, O2, are placed together in a container and allowed to react according to the following equation, which one, if any, is in excess and by how much? Also, how many moles of water, H2O, would be produced by the reaction?
To determine which reactant is in excess and by how much, we need to compare the molar ratio of the reactants to the stoichiometric ratio in the balanced chemical equation.
The balanced chemical equation for the reaction of hydrogen, H2, with oxygen, O2, to produce water, H2O, is:
2H2 + O2 -> 2H2O
From the balanced equation, we can see that 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water.
Given that we have 10.4 moles of hydrogen (H2) and 9.1 moles of oxygen (O2), we can calculate the moles of water produced by comparing the reactants to the stoichiometric ratio.
For hydrogen (H2):
- The stoichiometric ratio is 2 moles of H2 for every 1 mole of O2.
- Therefore, based on the given quantity of 9.1 moles of O2, the equivalent moles of H2 required would be (2/1) * 9.1 = 18.2 moles.
Comparing the required moles of H2 (18.2 moles) to the available moles (10.4 moles), we can see that the hydrogen is in excess by (18.2 - 10.4) = 7.8 moles.
Thus, hydrogen (H2) is in excess by 7.8 moles.
To calculate the moles of water (H2O) produced, we need to use the limiting reactant, which is oxygen (O2) in this case.
For oxygen (O2):
- The stoichiometric ratio is 1 mole of O2 for every 2 moles of H2O.
- Therefore, based on the given quantity of 9.1 moles of O2, the equivalent moles of H2O produced would be (2/1) * 9.1 = 18.2 moles.
Hence, if the reaction proceeds to completion, 18.2 moles of water (H2O) would be produced.