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How many moles of ammonia (NH3) can be produced from the reaction of 4.0 liters of hydrogen at 50.0°C and 1.2atm of pressure with excess nitrogen?

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Answer: Therefore, approximately 0.1247 moles of ammonia can be produced from the given reaction.

Step-by-step explanation:

To determine the number of moles of ammonia (NH3) produced from the given reaction, we need to use the ideal gas law and stoichiometry.

The balanced chemical equation for the reaction between hydrogen (H2) and nitrogen (N2) to form ammonia (NH3) is:

N2 + 3H2 → 2NH3

From the equation, we can see that three moles of hydrogen react with one mole of nitrogen to produce two moles of ammonia.

First, let's convert the given conditions of hydrogen to the appropriate units for the ideal gas law:

Volume of hydrogen = 4.0 liters

Temperature of hydrogen = 50.0°C = 50.0 + 273.15 = 323.15 K

Pressure of hydrogen = 1.2 atm

Now, let's calculate the number of moles of hydrogen using the ideal gas law equation:

PV = nRT

where:

P = pressure (in atm)

V = volume (in liters)

n = number of moles

R = gas constant (0.0821 L·atm/(mol·K))

T = temperature (in Kelvin)

n(H2) = PV / RT

n(H2) = (1.2 atm) * (4.0 L) / (0.0821 L·atm/(mol·K) * 323.15 K)

≈ 0.187 mol

Since the stoichiometry ratio is 3:2 (H2:NH3), we can conclude that 0.187 moles of hydrogen can produce (0.187/3) * 2 = 0.1247 moles of ammonia.

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