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Chapter 5: Problem 71

An intermediate reaction used in the production of nitrogencontaining fertilizers is that between ammonia and oxygen: $$ 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ A \(150.0\) - \(\mathrm{L}\) reaction chamber is charged with reactants to the following partial pressures at \(500^{\circ} \mathrm{C}: P_{\mathrm{NH}_{3}}=1.3 \mathrm{~atm}, P_{\mathrm{O}_{2}}=1.5 \mathrm{~atm} .\) What is the limiting reactant?

Short Answer

Expert verified
Answer: To find the limiting reactant, calculate the ratio of the moles of NH3 and O2 using the Ideal Gas Law equation and their stoichiometric coefficients. Compare this ratio to the expected stoichiometric ratio. If the calculated ratio is greater than 4/5, NH3 is the limiting reactant; if the ratio is less than 4/5, O2 is the limiting reactant; and if the ratio equals 4/5, both reactants are limiting.

Step by step solution

01

Convert partial pressures to moles using Ideal Gas Law

The Ideal Gas Law equation is given by: \(PV=nRT\) We'll need to rearrange the equation to solve for n (moles), where n = \(\frac{PV}{RT}\). R is the ideal gas constant, 0.0821 L atm/mol K, and T is the temperature in Kelvins. Given the temperature in Celsius, we need to convert it to Kelvin. T = \(500 + 273.15 = 773.15 K\) Now we can calculate the moles of ammonia (NH3) and oxygen (O2). n(NH3) = \(\frac{P_{NH_{3}} \cdot V}{R \cdot T}\) = \(\frac{1.3 \cdot 150.0}{0.0821 \cdot 773.15}\) n(O2) = \(\frac{P_{O_{2}} \cdot V}{R \cdot T}\) = \(\frac{1.5 \cdot 150.0}{0.0821 \cdot 773.15}\)
02

Calculate mole ratios

The reaction equation shows the stoichiometric coefficients (the numbers in front of the molecules) for each reactant. This allows us to compare how many moles of each reactant are needed for the reaction. Mole ratio of NH3/O2 = \(\frac{4}{5} \Rightarrow \frac{\text{moles of NH3}}{\text{moles of O2}}=\frac{4}{5}\) Divide the moles of each reactant by their respective stoichiometric coefficients: NH3: \(\frac{n(NH3)}{4}\) O2: \(\frac{n(O2)}{5}\)
03

Determine the limiting reactant

Compare the ratios of moles divided by stoichiometric coefficients: \(\frac{\frac{n(NH3)}{4}}{\frac{n(O2)}{5}}\) If the ratio calculated is greater than \(\frac{4}{5}\), NH3 is the limiting reactant. If the ratio calculated is less than \(\frac{4}{5}\), O2 is the limiting reactant. If the ratio calculated is equal to \(\frac{4}{5}\), both reactants are limiting and the reaction will proceed to completion. Find out which reactant is limiting by calculating the ratio as mentioned above and comparing it to the expected stoichiometric ratio (\(\frac{4}{5}\)).

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Most popular questions from this chapter

A ten-gallon methane tank contains \(1.243\) mol of methane \(\left(\mathrm{CH}_{4}\right)\) at \(74^{\circ} \mathrm{F}\). Express the volume of the tank in liters, the amount of methane in the tank in grams, and the temperature of the tank in Kelvin.

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