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Water can be added to magnesium nitride to create ammonia gas and magnesium oxide. If 2.0 moles of magnesium nitride is added to 4.0 moles of water, how many liters of ammonia gas should be produced if the gas is collected at 25.0 C and a pressure of 0.915 atm ?

Mg₃N₂​( s)+3H₂​O(l)→3MgO(s)+2NH₃​( g)

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Final answer:

To determine the volume of ammonia gas produced, use the ideal gas law equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. From the given amount of magnesium nitride, calculate the number of moles of ammonia gas produced. Then, use the ideal gas law to convert moles to volume.

Step-by-step explanation:

To determine the volume of ammonia gas produced, we need to use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to calculate the number of moles of ammonia gas produced. From the balanced chemical equation, we can see that 2 moles of ammonia gas are produced for every 2 moles of magnesium nitride.

Therefore, since we have 2.0 moles of magnesium nitride, we will also have 2.0 moles of ammonia gas.

Next, we need to convert the moles of ammonia gas to volume. We can rearrange the ideal gas law equation to solve for V: V = nRT/P.

Plugging in the values, we get V = (2.0 mol)(0.915 atm)(V)/(0.0821 L·atm/mol·K)(298.15 K).

Solving for V gives us V ≈ 38.8 L.

Therefore, approximately 38.8 liters of ammonia gas should be produced.

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