Answer :
Final answer:
In the given chemical reaction, Oxygen, O2, is the limiting reactant since it has fewer moles available for reaction. Using the stoichiometry and molar mass of nitrogen dioxide, NO2, we can calculate the resulting mass of NO2 to be approximately 283 grams.
Explanation:
In this Chemistry problem related to stoichiometry and limiting reactants, we need to consider the balanced chemical equation first: N2 + 2O2 → 2NO2. Molar mass of N2 is approximately 28 g/mol and for O2 it's 32 g/mol. Now, we need to convert mass into moles to determine the limiting reactant.
For Nitrogen, N2, moles = 105 g / 28 g/mol ≈ 3.75 moles.
For Oxygen, O2, moles = 98.5 g / 32 g/mol ≈ 3.08 moles.
From the balanced equation, it is clear that the molar ratio of N2:O2 is 1:2. Hence Oxygen O2 is the limiting reactant because we have fewer moles of it available for reaction. To calculate the number of grams of nitrogen dioxide, NO2, that can be formed, we multiply the moles of the limiting reactant by the stoichiometric factor and by the molar mass of NO2, which is about 46 g/mol. Therefore, mass of NO2 = 3.08 moles * 2 * 46 g/mol = 283 g approximately.
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