Question 12.25: Calculating Gaseous Product Volume from Gaseous Reactant Vol......
Calculating Gaseous Product Volume from Gaseous Reactant Volumes
How many liters of NO _2 gas at 3.00 atm pressure and 55°C can be produced from 20.0 L of N _2 gas at 1.00 atm pressure and 35°C and 30.0 L of O _2 gas at 2.00 atm pressure and 45°C, according to the following chemical reaction?
\textrm{N}_2 \textrm{(g) } + 2 \textrm{ O}_2 \textrm{(g) } \longrightarrow 2 \textrm{ NO}_2 \textrm{(g)}First, we must determine the limiting reactant because specific volumes of both reactants are given in the problem. This determination involves two separate volume-of-gas-A to moles-of-B calculations-one for each reactant. The pathway for these parallel calculations is
\boxed{\begin{matrix} \textrm{Volume} \\ \textrm{of gas A} \end{matrix}} \quad \underset{\textrm{volume}}{\underrightarrow{\textrm{Molar}}} \quad \boxed{\begin{matrix} \textrm{Moles} \\ \textrm{of A} \end{matrix}} \quad \underset{\textrm{coefficients}}{\underrightarrow{\textrm{Equation}}} \quad \boxed{\begin{matrix} \textrm{Moles} \\ \textrm{of B} \end{matrix}}The molar volume conversion factors for N _2 and O _2 will be different because the two gases are under different temperature-pressure conditions. Using the ideal gas law, we obtain the needed molar volume conversion factors as follows:
V_{\textrm{N}_2} = \frac{nRT}{P} = \frac{(1 \cancel{\textrm{mole}}) \Bigl(0.08206 \frac{\cancel{\textrm{atm}} \cdot \textrm{L}}{\cancel{\textrm{mole}} \cdot \cancel{\textrm{K}}} \Bigr) (308 \cancel{\textrm{K}})}{1.00 \cancel{\textrm{atm}}}= 25.27448 L (calculator answer)
= 25.3 L (correct answer)
= 13.04754 L (calculator answer)
= 13.0 L (correct answer)
With these two conversion factors known, the limiting reactant calculation proceeds in the normal manner.
For N _2 ,
20.0 \cancel{\textrm{L N}_2} \times \frac{1 \cancel{\textrm{mole N}_2}}{25.3 \cancel{\textrm{L N}_2}} \times \frac{2 \textrm{moles NO}_2}{1 \cancel{\textrm{mole N}_2}}= 1.5810276 moles NO _2 (calculator answer)
= 1.58 moles NO _2 (correct answer)
For O _2 ,
30.0 \cancel{\textrm{L O}_2} \times \frac{1 \cancel{\textrm{mole O}_2}}{13.0 \cancel{\textrm{L O}_2}} \times \frac{2 \textrm{moles NO}_2}{2 \cancel{\textrm{mole O}_2}}= 2.3076923 moles NO _2 (calculator answer)
= 2.31 moles NO _2 (correct answer)
Nitrogen (N _2 ) is the limiting reactant since it produces fewer moles of NO _2 .
The overall calculation is a volume-of-gas-A to volume-of-gas-B calculation.
In doing the limiting reactant calculation we obtained moles of B (1.58 moles of NO _2 ). Thus, all that is left to do is to go from moles-of-B to volume-of-gas-B. A new molar volume conversion factor is needed since NO _2 is present at different temperature-pressure conditions than was either N _2 or O _2 .
This molar volume conversion factor is
V_{\textrm{NO}_2} = \frac{nRT}{P} = \frac{(1 \cancel{\textrm{mole}}) \Bigl(0.08206 \frac{\cancel{\textrm{atm}} \cdot \textrm{L}}{\cancel{\textrm{mole}} \cdot \cancel{\textrm{K}}} \Bigr) (328 \cancel{\textrm{K}})}{3.00 \cancel{\textrm{atm}}}= 8.9718933 L NO _2 (calculator answer)
= 8.97 L NO _2 (correct answer)
The moles-of-B to grams-of-B calculation, a one-step calculation, becomes
1.58 \cancel{\textrm{moles NO}_2} \times \frac{8.97 \textrm{L NO}_2}{1 \cancel{\textrm{mole NO}_2}}= 14.1726 L NO _2 (calculator answer)
= 14.2 L NO _2 (correct answer)