ammoniagas and oxygengas react to form nitrogen gas and water vapor. suppose you have of and of in a reactor. calculate the largest amount of that could be produced. round your answer to the nearest .

Answer:

Explanation:

To calculate the largest amount of nitrogen gas (N2) that could be produced when ammonia gas (NH3) and oxygen gas (O2) react, you need to consider the balanced chemical equation for the reaction:

4 NH3(g) + 5 O2(g) -> 4 N2(g) + 6 H2O(g)

From the balanced equation, you can see that 4 moles of NH3 produce 4 moles of N2.

Now, let's calculate the number of moles of NH3 and O2 you have:

Moles of NH3 = 4.0 grams / molar mass of NH3

Moles of O2 = 8.0 grams / molar mass of O2

The molar mass of NH3 (ammonia) is approximately 17.03 g/mol, and the molar mass of O2 (oxygen) is approximately 32.00 g/mol.

Moles of NH3 = 4.0 g / 17.03 g/mol ≈ 0.235 moles

Moles of O2 = 8.0 g / 32.00 g/mol ≈ 0.250 moles

Now, you need to determine the limiting reactant, which is the reactant that is completely consumed and limits the amount of product formed. To do this, compare the mole ratio from the balanced equation to the actual moles of NH3 and O2:

NH3 : O2 = 4 moles : 5 moles

Since you have fewer moles of NH3 (0.235 moles) than the required 0.25 moles based on the mole ratio, NH3 is the limiting reactant.

Now, calculate the moles of N2 produced using the mole ratio:

Moles of N2 = Moles of NH3 (since it's the limiting reactant)

Moles of N2 = 0.235 moles

Now, you can convert moles of N2 to grams:

Mass of N2 = Moles of N2 x Molar mass of N2

The molar mass of N2 (nitrogen gas) is approximately 28.02 g/mol.

Mass of N2 = 0.235 moles x 28.02 g/mol ≈ 6.59 grams

Rounding to the nearest gram, the largest amount of N2 that could be produced is approximately 7 grams.