Question 5.SP.24: Using the balanced equation, N2(g)+ O2(g)→2NO(g), calculate ...
Using the balanced equation, N_{2}(g) + O_{2}(g) \longrightarrow 2NO(g), calculate the number of grams of NO (molar mass 30.01 g/mol) formed 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).
Three steps are required:
• Determine the limiting reactant.
• Use the mole ratio from the balanced equation to calculate the number of moles of product formed.
• Convert the moles of product to grams using the molar mass.
[1] Determine the limiting reactant.
• Convert the number of grams of each reactant to the number of moles using the molar masses.
• Use a mole–mole conversion factor to determine which reactant is present in excess.
• As shown in Sample Problem 5.23, when 10.0 g of N_{2} (0.357 mol) react with 10.0 g (0.313 mol)
of O_{2}, O_{2} is the limiting reactant.
[2] Convert the number of moles of limiting reactant to the number of moles of product using a mole–mole conversion factor.
\begin{matrix} & & \text{ mole–mole conversion factor}& &&\\ \\ 0.313 \cancel{ mol } O_{2}&\times &\frac{\text{ 2 mol } NO }{1 \cancel{ mol } O_{2}} &= & 0.626 \text{ mol } NO \text{ formed } \end{matrix}
[3] Convert the number of moles of product to the number of grams using the molar mass.
\begin{matrix} 0.626 \cancel{ mol } NO &= &\frac{\text{30.01 g } NO}{1 \cancel{ mol } NO} &= & 18.8 \text{ g } NO \text{ formed } \\&&&&Answer \end{matrix}