In the Haber process, nitrogen reacts with hydrogen to produce ammonia: \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)\) (a) Suppose \(2.0 \mathrm{~L}\) of \(\mathrm{N}_{2}\) gas at STP is combined with \(6.0 \mathrm{~L}\) of \(\mathrm{H}_{2}\) gas, with the two gases being at the same temperature and pressure. Is this reaction being run in a balanced fashion or in a limiting fashion? Explain how you can tell without doing any calculations. (b) If \(50.0 \mathrm{~L}\) of \(\mathrm{N}_{2}\) gas at \(200.0 \mathrm{lb} / \mathrm{in}^{2}\) and \(22.0^{\circ} \mathrm{C}\) is combined with \(100.0 \mathrm{~L}\) of \(\mathrm{H}_{2}\) gas at \(240.0 \mathrm{lb} / \mathrm{in}^{2}\) and \(22.0^{\circ} \mathrm{C}\), what mass in grams of ammonia is produced? \(\left[14.70 \mathrm{lb} / \mathrm{in}^{2}=760.0 \mathrm{~mm} \mathrm{Hg}\right]\)

Step by step solution

01

Reaction Analysis

In the Haber process, the balanced reaction between nitrogen (N2) and hydrogen (H2) forming ammonia (NH3) is given by: \(N_{2}(g) + 3H_{2}(g) \rightarrow 2NH_{3}(g)\).This reaction shows that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3. Part (a)

02

Compare the Reaction Mole Ratios

If we have 2.0 L of N2 gas and 6.0 L of H2 gas at the same temperature and pressure, let's compare their mole ratios with the stoichiometry of the given reaction. Since Avogadro's law states that equal volumes of gases at the same pressure and temperature contain equal numbers of moles, we can simply compare the volume ratios, which are: \[\frac{N_{2}}{H_{2}} = \frac{2.0 L}{6.0 L} = \frac{1}{3}\] This ratio exactly matches the stoichiometry of the balanced reaction (\(N_{2}:H_{2} = 1:3\)), so it can be concluded that this reaction is being run in a balanced fashion. Part (b)

03

Convert Pressure Units

First, we need to convert the pressure of both gases into the same unit. Let's use mmHg according to the given conversion factor: \[N_{2}: \frac{200.0 lb/in^{2}}{1} \times \frac{760.0 mmHg}{14.7 lb/in^{2}} = 1024.5 mmHg\] \[H_{2}: \frac{240.0 lb/in^{2}}{1} \times \frac{760.0 mmHg}{14.7 lb/in^{2}} = 1229.3 mmHg\]

04

Calculate Moles of Each Gas

Now, we can use the ideal gas law to calculate moles of N2 and H2: \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant (\(62.4 \frac{mmHg \cdot L}{mol \cdot K}\)), and \(T\) is the temperature in Kelvin (\(22.0^{\circ}C + 273.15 = 295.15 K\)). \[n_{N_{2}} = \frac{P_{N_{2}}V_{N_{2}}}{RT} = \frac{ (1024.5 mmHg)(50.0 L)}{(62.4 \frac{mmHg \cdot L}{mol \cdot K})(295.15 K)} = 11.05 mol\] \[n_{H_{2}} = \frac{P_{H_{2}}V_{H_{2}}}{RT} = \frac{ (1229.3 mmHg)(100.0 L)}{(62.4 \frac{mmHg \cdot L}{mol \cdot K})(295.15 K)} = 28.01 mol\]

05

Determine the Limiting Reactant

Now let's find out the limiting reactant by comparing the mole ratios of the two gases: \[\frac{n_{N_{2}}}{n_{H_{2}}} = \frac{11.05 mol}{28.01 mol} = 1:2.54\] This mole ratio is different from the stoichiometry of the balanced reaction (\(1:3\)), indicating that N2 is in excess, and H2 is the limiting reactant.

06

Calculate Moles and Mass of Ammonia

From the balanced equation, we know that 3 moles of H2 produce 2 moles of NH3. Therefore, the moles of NH3 produced are: \[n_{NH_{3}} = \frac{2}{3} n_{H_{2}} = \frac{2}{3} \times 28.01 mol = 18.67 mol\] Finally, we can find the mass of ammonia produced by using its molar mass (\(M_{NH_{3}} = 14.01 + 3(1.01) = 17.03 g/mol\)): \[m_{NH_{3}} = n_{NH_{3}} \times M_{NH_{3}} = 18.67 mol \times 17.03 g/mol = 317.88 g\] The mass of ammonia produced is approximately \(317.88 g\).

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