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The reaction shown represents the oxidation of ammonia:

[tex]\[ 4 NH_3(aq) + 5 O_2(g) \rightarrow 4 NO(g) + 6 H_2O(l) \][/tex]

How many grams of water [tex]\((H_2O)\)[/tex] will be formed when 22 grams of oxygen [tex]\((O_2)\)[/tex] reacts with an excess of ammonia [tex]\((NH_3)\)[/tex]?

A. 37.1 grams
B. 58.2 grams
C. 87.4 grams
D. 98.3 grams

Answer :

Sure! Let's solve this problem step-by-step:

We have the chemical reaction:

[tex]\[ 4 \, \text{NH}_3 + 5 \, \text{O}_2 \rightarrow 4 \, \text{NO} + 6 \, \text{H}_2\text{O} \][/tex]

We need to determine how many grams of water ([tex]\(\text{H}_2\text{O}\)[/tex]) will be formed when 22 grams of oxygen ([tex]\(\text{O}_2\)[/tex]) react.

### Step 1: Determine Molar Mass

1. Molar mass of [tex]\(\text{O}_2\)[/tex]:
- Oxygen ([tex]\(\text{O}\)[/tex]) has a molar mass of approximately 16.00 g/mol.
- Therefore, [tex]\(\text{O}_2\)[/tex] has a molar mass of [tex]\(32.00 \, \text{g/mol}\)[/tex].

2. Molar mass of [tex]\(\text{H}_2\text{O}\)[/tex]:
- Hydrogen ([tex]\(\text{H}\)[/tex]) has a molar mass of approximately 1.01 g/mol.
- Therefore, [tex]\(\text{H}_2\)[/tex] has a molar mass of [tex]\(2.02 \, \text{g/mol}\)[/tex].
- Add oxygen’s molar mass, so [tex]\(\text{H}_2\text{O}\)[/tex] has a molar mass of [tex]\(18.02 \, \text{g/mol}\)[/tex].

### Step 2: Calculate Moles of Oxygen

We have 22 grams of [tex]\(\text{O}_2\)[/tex].

- Calculate moles of [tex]\(\text{O}_2\)[/tex]:
[tex]\[
\text{Moles of } \text{O}_2 = \frac{\text{Mass of } \text{O}_2}{\text{Molar mass of } \text{O}_2} = \frac{22 \, \text{g}}{32.00 \, \text{g/mol}} = 0.6875 \, \text{moles}
\][/tex]

### Step 3: Use Stoichiometry to Find Moles of Water

According to the balanced equation, 5 moles of [tex]\(\text{O}_2\)[/tex] produce 6 moles of [tex]\(\text{H}_2\text{O}\)[/tex].

- Since we have 0.6875 moles of [tex]\(\text{O}_2\)[/tex], calculate the moles of [tex]\(\text{H}_2\text{O}\)[/tex]:
[tex]\[
\text{Moles of } \text{H}_2\text{O} = \left(\frac{6}{5}\right) \times 0.6875 = 0.825 \, \text{moles}
\][/tex]

### Step 4: Calculate Mass of Water

- Calculate the mass of [tex]\(\text{H}_2\text{O}\)[/tex]:
[tex]\[
\text{Mass of } \text{H}_2\text{O} = \text{Moles of } \text{H}_2\text{O} \times \text{Molar mass of } \text{H}_2\text{O} = 0.825 \times 18.02 \, \text{g/mol} = 14.87 \, \text{grams}
\][/tex]

Therefore, when 22 grams of oxygen react, 14.87 grams of water will be formed.

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