16.0 grams of H2 react with 64.0 grams of O2 according to the following equation:2H2(g) + O2(g) → 2H2O(l)Determine which of the following species will be present in the reaction mixture at the end of the reaction. MH = 1.01 g mol-1MO = 16.00 g mol-1H2, H2O, and O2O2, H2OH2, O2H2, H2O
Question
16.0 grams of H2 react with 64.0 grams of O2 according to the following equation:2H2(g) + O2(g) → 2H2O(l)Determine which of the following species will be present in the reaction mixture at the end of the reaction. MH = 1.01 g mol-1MO = 16.00 g mol-1H2, H2O, and O2O2, H2OH2, O2H2, H2O
Solution
To determine which species will be present at the end of the reaction, we first need to identify the limiting reactant, as it will be completely consumed in the reaction.
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Calculate the number of moles for each reactant:
For H2: 16.0 g / 2.02 g/mol = 7.92 mol For O2: 64.0 g / 32.00 g/mol = 2.00 mol
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Compare the mole ratio:
The balanced chemical equation tells us that the reaction consumes H2 and O2 in a 2:1 ratio. However, we have more than that ratio in our reactants (7.92 mol H2 : 2.00 mol O2). This means that O2 is the limiting reactant.
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Determine the remaining species:
Since O2 is the limiting reactant, it will be completely consumed in the reaction. H2 will react with all the O2 to produce H2O, but there will be some H2 left over because we started with more than the necessary amount.
Therefore, the species present at the end of the reaction will be H2 and H2O.
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