Consider the reaction. mc015-1.jpg How many grams of N2 are required to produce 100.0 L of NH3 at STP?

Answers

Answer 1

Answer: The grams of N2 that are required to produce 100.0 l of NH3 at STP

At stp 1moles = 22.4 l. what about 100.0 L of NH3

= 100 / 22.4 lx1 moles = 4.46 moles of NH3

write the reacting equation

N2+3H2 =2NH3
by use of mole ratio between N2 to NH3 which is 1:2 the moles of N2 =4.46/2 =2.23 moles of N2

mass = moles x molar mass

= 2.23moles x 28 g/mol = 62.4 grams

Answer 2

Answer:

Final answer:

To produce 100.0 L of NH3 at STP, 62.4 grams of N2 are required.

Explanation:

The balanced equation for the reaction is:

N2(g) + 3H2(g) → 2NH3(g)

According to the balanced equation, one mole of N2 reacts with three moles of H2 to produce two moles of NH3. From this information, we can use stoichiometry to determine the mass of N2 required to produce 100.0 L of NH3 at STP.

First, we need to convert liters of NH3 to moles using the ideal gas law and the molar volume of a gas at STP (~22.4 L/mol). Once we have the moles of NH3, we can use the mole ratio from the balanced equation to calculate the moles of N2. Finally, we can use the molar mass of N2 to convert moles to grams.

Let's calculate:

Convert liters of NH3 to moles of NH3: 100.0 L * (1 mol/22.4 L) = 4.46 mol NH3
Calculate moles of N2 using the mole ratio: 4.46 mol NH3 * (1 mol N2/2 mol NH3) = 2.23 mol N2
Convert moles of N2 to grams of N2 using the molar mass of N2 (28 g/mol): 2.23 mol N2 * 28 g/mol = 62.4 g N2

Therefore, 62.4 grams of N2 are required to produce 100.0 liters of NH3 at STP.

Learn more about Stoichiometry of Gaseous Substances here:

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