Question

The reaction shown represents the oxidation of ammonia NH3.

4NH3 + 5O2 --> 4NO + 6H2O

How many grams of water will be formed when 2.0 moles of ammonia reacts with excess oxygen O2 ? The molar mass of water is 18 g/mol. The molar mass of Ammonia is 17.04 g/mol.

Responses

54 grams
54 grams

64 grams
64 grams

110 grams
110 grams

51 grams

Answer 1

To determine how many grams of water will be formed when 2.0 moles of ammonia (NH₃) reacts with excess oxygen (O₂), we can follow these steps:

1. **Use the balanced chemical equation:**
\[
4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2\text{O}
\]

2. **Determine the stoichiometry from the equation:**
According to the equation, 4 moles of NH₃ produce 6 moles of H₂O.

3. **Find how many moles of water are produced from 2.0 moles of NH₃:**
We can set up a ratio:
\[
\frac{6 \text{ moles H}_2O}{4 \text{ moles NH}_3} = \frac{x \text{ moles H}_2O}{2.0 \text{ moles NH}_3}
\]
Cross-multiplying gives:
\[
x = 2.0 \times \frac{6}{4} = 2.0 \times 1.5 = 3.0 \text{ moles of H}_2O
\]

4. **Convert moles of water to grams:**
The molar mass of water (H₂O) is 18 g/mol. Thus:
\[
\text{Mass of H}_2O = \text{moles} \times \text{molar mass} = 3.0 \text{ moles} \times 18 \text{ g/mol} = 54 \text{ grams}
\]

Therefore, when 2.0 moles of ammonia reacts with excess oxygen, 54 grams of water will be formed.

The correct response is **54 grams**.