Question

How many moles of N2 are produced when 20 grams of NH3 react

with 25 grams of O2?
4NH3(g) + 3O2(g) ——-> 2N2(g) + 6H2O(g)
a) 0.78
b) 0.59
c) 1.2
d) 0.52

Answer 1

The correct number of moles of N₂ produced when 20 grams of NH₃ react with 25 grams of O₂, given that O₂ is the limiting reactant, is 0.52 moles.

Step-by-step explanation:

To solve this problem, we need to determine the limiting reactant and then use stoichiometry to find the number of moles of N₂ produced. The balanced chemical equation is 4NH₃(g) + 3O₂(g) → 2N₂(g) + 6H₂O(g). We will calculate the number of moles of both NH₃ and O₂ and use the mole ratio from the equation to identify the limiting reactant.

First, we calculate the number of moles of NH₃ using its molar mass (17 g/mol):
17 g/mol NH₃ * X mol = 20 g NH₃
X = 20 g / 17 g/mol
X = 1.18 mol NH₃

Now, we calculate the number of moles of O₂ using its molar mass (32 g/mol):
32 g/mol O₂ * Y mol = 25 g O₂
Y = 25 g / 32 g/mol
Y = 0.78 mol O₂

According to the balanced equation, the molar ratio of NH₃ to O₂ is 4:3. Therefore, the number of moles of O₂ needed to react with 1.18 mol of NH₃ is (1.18 mol NH₃ * 3 mol O₂) / 4 mol NH₃ = 0.885 mol O₂. Since we have only 0.78 mol O₂ available, O₂ is the limiting reactant.

Now we use the molar ratio of O₂ to N₂ from the balanced equation to find the amount of N₂ produced from 0.78 mol of O₂: (0.78 mol O₂ * 2 mol N₂) / 3 mol O₂ = 0.52 mol N₂.

The correct answer is therefore (d) 0.52.