Questions: Determine the limiting reactant and calculate the number of grams of nitrogen dioxide, NO2, that can be formed when 105 g of nitrogen, N2, reacts with 98.5 g of oxygen, O2. N2(g) + 2 O2(g) → 2 NO2(g)

Solution Steps

First, we need to calculate the number of moles of each reactant. The molar mass of nitrogen (\(\mathrm{N}_2\)) is approximately 28.02 g/mol, and the molar mass of oxygen (\(\mathrm{O}_2\)) is approximately 32.00 g/mol.

For nitrogen: \[ \text{Moles of } \mathrm{N}_2 = \frac{105 \, \text{g}}{28.02 \, \text{g/mol}} \approx 3.748 \, \text{mol} \]

For oxygen: \[ \text{Moles of } \mathrm{O}_2 = \frac{98.5 \, \text{g}}{32.00 \, \text{g/mol}} \approx 3.078 \, \text{mol} \]

The balanced chemical equation is: \[ \mathrm{N}_{2}(g) + 2 \mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NO}_{2}(g) \]

According to the stoichiometry of the reaction, 1 mole of \(\mathrm{N}_2\) reacts with 2 moles of \(\mathrm{O}_2\). Therefore, we need to check which reactant is limiting.

Calculate the required moles of \(\mathrm{O}_2\) for the available \(\mathrm{N}_2\): \[ \text{Required moles of } \mathrm{O}_2 = 3.748 \, \text{mol} \times 2 = 7.496 \, \text{mol} \]

Since we only have 3.078 moles of \(\mathrm{O}_2\), oxygen is the limiting reactant.

Since \(\mathrm{O}_2\) is the limiting reactant, we use its moles to calculate the moles of \(\mathrm{NO}_2\) produced. According to the balanced equation, 2 moles of \(\mathrm{O}_2\) produce 2 moles of \(\mathrm{NO}_2\).

\[ \text{Moles of } \mathrm{NO}_2 = 3.078 \, \text{mol} \times \frac{2 \, \text{mol} \, \mathrm{NO}_2}{2 \, \text{mol} \, \mathrm{O}_2} = 3.078 \, \text{mol} \]

Now, calculate the mass of \(\mathrm{NO}_2\) produced. The molar mass of \(\mathrm{NO}_2\) is approximately 46.01 g/mol.

\[ \text{Mass of } \mathrm{NO}_2 = 3.078 \, \text{mol} \times 46.01 \, \text{g/mol} \approx 141.5 \, \text{g} \]

Final Answer