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At STP, if 35.8 g of NH\(_3\) is reacted with an excess of nitrogen monoxide, how many liters of N\(_2\) can be produced?

Reaction: \( \text{NH}_3 + \text{NO} \rightarrow \text{N}_2 + \text{H}_2\text{O} \)

Answer :

Final answer:

To calculate the volume of N₂ produced when NH3 is reacted with NO at STP, we convert the mass of NH3 to moles and use the ideal gas law equation. The volume of N₂ produced is 2.10 L.


Explanation:

To find the number of liters of N₂ produced, we first need to convert the mass of NH3 to moles. Using the molar mass of NH3 (17.03 g/mol), we have 35.8 g ÷ 17.03 g/mol = 2.10 mol NH3. From the balanced equation, we can see that the ratio between NH3 and N₂ is 1:1. So, 2.10 mol of NH3 will produce 2.10 mol of N₂.

Next, we can use the ideal gas law to calculate the volume of N₂. The ideal gas law equation is PV = nRT, where P represents pressure, V represents volume, n represents the number of moles, R represents the gas constant, and T represents temperature. Since the temperature and pressure are at STP (standard temperature and pressure), we can use specific values for these. STP is defined as 273.15 K and 1 atm.

Let's assume that the volume of N₂ is V. From the equation, we can set up the following proportion: (2.10 mol / V) = (1 mol / V). Solving for V, we find that the volume of N₂ produced is 2.10 L.


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