Question 5.SP.23: Using the balanced equation, N2(g) + O2(g) → 2NO(g), determi...
Using the balanced equation, N_{2}(g) + O_{2}( g) \longrightarrow 2NO( g) , determine the limiting reactant when 10.0 g of N_{2} (molar mass 28.02 g/mol) react with 10.0 g of O_{2} (molar mass 32.00 g/mol).
Convert the number of grams to the number of moles for each reactant, and then use the steps in Sample Problem 5.21 to determine the limiting reactant.
[1] Convert the number of grams of each reactant to the number of moles using the molar masses.
\begin{matrix} \text{Moles of }N_{2} : & & 10.0 \cancel{g}\enspace N_{2} & \times & \frac{\text{1 mol}\enspace N_{2}}{28.02 \cancel{g} \enspace N_{2}} &=& \text{0.357 mol }N_{2}\\ \\ \text{Moles of }O_{2} : & & 10.0 \cancel{g}\enspace O_{2} & \times & \frac{\text{1 mol}\enspace O_{2}}{32.00 \cancel{g} \enspace O_{2}} &=& \text{0.313 mol }O_{2} \end{matrix}
[2] Determine the limiting reactant as in Sample Problem 5.21.
• Identify N_{2} as the original quantity and use the mole ratio from the alanced equation as a conversion factor to calculate how much O_{2} is needed for complete reaction.
• Since the amount of O_{2} present is less than what is needed, O_{2} is the limiting reactant.