<h3>
Answer:</h3>
78.34 g
<h3>
Explanation:</h3>
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.
<h3>Step 1: Writing the balanced equation for the reaction </h3>
The Balanced equation for the reaction is;
N₂(g) + 3H₂(g) → 2NH₃(g)
<h3>Step 2: Calculating the number of moles of NH₃</h3>
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
<h3>Step 3: Calculating the mass of ammonia produced </h3>
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
I think it’s A but I’m not sure. Sorry if I’m wrong
Answer:
take these as notes ig:)
Explanation:
reaction equations to make connections between the reactants and products in reactions. ... pathway: mol of O2 → mol of H2O → grams of H2O. 3.5 mol ... 25 g KClO3 ... How many grams of hydrogen gas are produced from the reaction of 221.
