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Answer :
Final answer:
24.9 L of N2 produces 49.8 L of NH3. Using the Ideal Gas Law, this volume of NH3 corresponds to about 2.02 moles. With a molar mass of 17 g/mol, this amounts to approximately 34.34 grams of NH3.
Explanation:
To solve this problem, we need to use the Ideal Gas Law (PV = nRT) and the balanced chemical equation for the process: N2 + 3H2 → 2NH3. According to this equation, the mole ratio between N2 and NH3 is 1:2. This means that for every 1 mole (or in this case, 24.9 L) of N2 reacted, 2 moles (or 49.8 L, because we assume standard temperature and pressure where 1 mole = 22.4 L) of NH3 are produced.
To find the mass of this NH3, we first need to find the number of moles. Using the Ideal Gas Law, we can calculate the number of moles as n = PV/RT. Given P = 97.8 kPa, V = 49.8 L, R = 8.31 (kPa L)/(K mol), and T = 23.7°C + 273.15 = 296.85 K, we find n ≈ 2.02 moles of NH3.
Finally, the molar mass of NH3 is approximately 17 g/mol, so the mass of the produced NH3 can be calculated as 2.02 moles * 17 g/mol ≈ 34.34 grams.
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Answer:
A. 49.8L of NH3.
B. 33.83g.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is shown below:
N2 + 3H2 —> 2NH3
A. From the balanced equation above,
1L of N2 produced 2L of NH3.
Therefore, 24.9L of N2 will produce = 24.9 x 2 = 49.8L of NH3.
Therefore, 49.8L of NH3 is produced from the reaction.
B. Determination of the mass NH3 produced.
First, we shall determine the number of mole of NH3 produced. This can be obtained as follow:
Volume (V) = 49.8L
Pressure (P) = 97.8 kPa = 97.8/101.325 = 0.97atm
Temperature (T) = 23.7°C = 23.7°C + 273 = 296.7K
Gas constant (R) = 0.082atm.L/Kmol
Number of mole (n) =...?
PV = nRT
n = PV /RT
n = (0.97 x 49.8) / (0.082 x 296.3)
n = 1.99 mole
Next, we shall convert 1.99 mole to NH3 to grams. This is illustrated below:
Number of mole NH3 = 1.99 mole
Molar mass of NH3 = 14 + (3x1) = 17g/mol
Mass of NH3 =..?
Mass = mole x molar Mass
Mass of NH3 = 1.99 x 17
Mass of NH3 = 33.83g
Therefore, 33.83g of NH3 is produced.
