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If 13.5 L of nitrogen gas reacts with 17.8 L of hydrogen gas at STP, according to the following reaction, what mass of ammonia (NH3) would be produced in the given reaction?

N2 + 3 H2 → 2 NH3

User Ragulka
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Final answer:

To determine the mass of ammonia produced, we need to use the given volumes of nitrogen and hydrogen gas and convert them to moles. Then, using the stoichiometric ratio from the balanced equation, we can calculate the moles of ammonia produced. Finally, we can convert the moles of ammonia to grams using the molar mass of ammonia.

Step-by-step explanation:

To determine the mass of ammonia produced, we need to use the given volumes of nitrogen and hydrogen gas and convert them to moles. From the balanced equation, we can see that 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia. Using the molar volume of gas at STP (22.4 L/mol), we can calculate the number of moles of nitrogen and hydrogen. Then, using the molar mass of ammonia, we can calculate the mass of ammonia produced.

First, let's convert the given volumes of nitrogen and hydrogen gas to moles:

Moles of nitrogen gas = 13.5 L / 22.4 L/mol = 0.604 moles

Moles of hydrogen gas = 17.8 L / 22.4 L/mol = 0.795 moles

Next, we can determine the limiting reagent, which is the reactant that is completely consumed in the reaction. To do this, we compare the moles of nitrogen and hydrogen to the stoichiometric ratio in the balanced equation. Since the ratio is 1:3 (nitrogen to hydrogen), we can see that the nitrogen is the limiting reagent.

Now, using the stoichiometric ratio, we can calculate the moles of ammonia produced:

Moles of ammonia = 0.604 moles of nitrogen x (2 moles of ammonia / 1 mole of nitrogen) = 1.208 moles

Finally, we can convert the moles of ammonia to grams using the molar mass of ammonia (17.03 g/mol):

Mass of ammonia = 1.208 moles x 17.03 g/mol = 20.57 grams

Therefore, the mass of ammonia produced in the given reaction is approximately 20.57 grams.

User KDrewiske
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