Question

If 15 g of N _2and 12.5 g of H.._2 are reacted, what mass of NH _3

​ (theoretical yield) is produced?
A) 17.5 g
B) 20 g
C) 22.5 g
D) 25 g

Answer 1

The theoretical yield of NH3 produced from 15 g of N2 and 12.5 g of H2 is calculated to be 18.22 g, but when rounding to the nearest available choice, the correct answer is (A) 17.5 g.

Step-by-step explanation:

To determine the theoretical yield of NH3 produced from 15 g of N2 and 12.5 g of H2, we use the chemical equation N2(g) + 3H2(g) → 2NH3(g).

First, we need to calculate the moles of N2 and H2 using their molar masses (N2: 28.02 g/mol, H2: 2.02 g/mol).

This gives us 15 g N2 / 28.02 g/mol ≈ 0.535 moles of N2 and 12.5 g H2 / 2.02 g/mol ≈ 6.188 moles of H2.

According to the balanced equation, 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

The limiting reagent is N2 because it will run out first. Therefore, 0.535 moles of N2 × 2 moles NH3 / 1 mole N2 gives us 1.070 moles of NH3.

Finally, converting moles of NH3 to grams (NH3: 17.03 g/mol), we have 1.070 moles × 17.03 g/mol = 18.22 g of NH3.

However, since the options provided do not include this value, rounding to the nearest available choice gives us 17.5 g of NH3.

So, the correct answer is (A) 17.5 g.