How many grams of NH3 can be produced from 2.30 mol of N2 and excess H2.

78.34 g

Explanation:

From the question we are given;

Moles of Nitrogen gas as 2.3 moles

we are required to calculate the mass of NH₃ that may be reproduced.

Step 1: Writing the balanced equation for the reaction

The Balanced equation for the reaction is;

N₂ (g) + 3H₂ (g) → 2NH₃ (g)

Step 2: Calculating the number of moles of NH₃

From the equation 1 mole of nitrogen gas reacts to produce 2 moles of NH₃

Therefore, the mole ratio of N₂ to NH₃ is 1 : 2

Thus, Moles of NH₃ = Moles of N₂ * 2

= 2.3 moles * 2

= 4.6 moles

Step 3: Calculating the mass of ammonia produced

Mass = Moles * molar mass

Molar mass of ammonia gas = 17.031 g/mol

Therefore;

Mass = 4.6 moles * 17.031 g/mol

= 78.3426 g

= 78.34 g

Thus, the mass of NH₃ produced is 78.34 g