Answer :
Final answer:
The limiting reactant is N2. The mass of NO2 produced is 345.04 g.
Explanation:
To determine the limiting reactant and mass of $ ext{NO}_2$, we first need to write a balanced chemical equation for the reaction between $ ext{N}_2$ and $ ext{O}_2$. The balanced equation is:
$$ ext{N}_2 + ext{O}_2
ightarrow 2 ext{NO}_2$$
Next, we calculate the moles of $ ext{N}_2$ and $ ext{O}_2$ using their given masses and molar masses:
$$ ext{moles of N}_2 = rac{105 ext{ g}}{28.013 ext{ g/mol}} = 3.75 ext{ mol}$$
$$ ext{moles of O}_2 = rac{98.5 ext{ g}}{31.9988 ext{ g/mol}} = 3.08 ext{ mol}$$
Based on the balanced equation, we can see that $ ext{N}_2$ and $ ext{O}_2$ have a 1:1 molar ratio. Therefore, $ ext{N}_2$ is the limiting reactant because it will be completely consumed before all of the $ ext{O}_2$ reacts. The mass of $ ext{NO}_2$ produced can be calculated using the stoichiometry of the reaction:
$$ ext{moles of NO}_2 = 2 imes ext{moles of N}_2 = 2 imes 3.75 ext{ mol} = 7.5 ext{ mol}$$
$$ ext{mass of NO}_2 = ext{moles of NO}_2 imes ext{molar mass of NO}_2$$
$$ ext{mass of NO}_2 = 7.5 ext{ mol} imes 46.0055 ext{ g/mol} = 345.04 ext{ g}$$
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