The molar mass of NH3 is 17.03 g/mol, and the molar mass of H2O is 18.02 g/mol. Consider the following (already balanced) chemical equation: 4 NH3(g) + 7 O2(g) ? 4 NO2(g) + 6 H2O(g) How many moles of H2O are formed when 58.6 g of NH3 react with excess oxygen according to the above equation?

1.50 x 10
2.29
3.44
87.9
5.16

The correct answer and explanation is :

Let’s solve it carefully first:

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Given:

• Molar mass of NH₃ = 17.03 g/mol
• Balanced reaction:
  4 NH₃(g) + 7 O₂(g) → 4 NO₂(g) + 6 H₂O(g)
• Mass of NH₃ = 58.6 g

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Step 1: Find moles of NH₃.
[
\text{moles of NH}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{58.6\ \text{g}}{17.03\ \text{g/mol}} = 3.44\ \text{mol}
]

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Step 2: Use the mole ratio from the balanced equation.
From the reaction,
4 moles of NH₃ produce 6 moles of H₂O.

Set up the proportion:
[
\text{moles of H}_2\text{O} = 3.44\ \text{mol NH}_3 \times \frac{6\ \text{mol H}_2\text{O}}{4\ \text{mol NH}_3}
]
[
= 3.44 \times 1.5
]
[
= 5.16\ \text{mol H}_2\text{O}
]

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✅ Final Answer: 5.16

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Explanation (Approx. 300 words):

In this question, we are asked to find out how many moles of water (H₂O) are produced when a known mass of ammonia (NH₃) reacts completely with excess oxygen. The first step is to convert the mass of NH₃ into moles, because chemical equations work in terms of moles, not grams.

The molar mass of NH₃ is given as 17.03 g/mol. Using the basic formula for moles, which is mass divided by molar mass, we find that:
[
\text{moles of NH}_3 = \frac{58.6\ \text{g}}{17.03\ \text{g/mol}} = 3.44\ \text{mol}
]

Next, we refer to the balanced chemical equation:
[
4\ \text{NH}_3 + 7\ \text{O}_2 \rightarrow 4\ \text{NO}_2 + 6\ \text{H}_2\text{O}
]
This equation tells us that 4 moles of NH₃ yield 6 moles of H₂O. Thus, for every mole of NH₃, 1.5 moles of H₂O are formed.

Multiplying the moles of NH₃ (3.44 mol) by the ratio (1.5), we get:
[
3.44\ \text{mol} \times 1.5 = 5.16\ \text{mol H}_2\text{O}
]

This calculation assumes that oxygen is in excess, so NH₃ is the limiting reactant.

Finally, we conclude that 5.16 moles of water are produced when 58.6 grams of ammonia react under these conditions.

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