The reaction shown represents the oxidation of ammonia by oxygen.

[tex]\[4 \text{NH}_3(aq) + 5 \text{O}_2(g) \rightarrow 4 \text{NO}(g) + 6 \text{H}_2\text{O}(l)\][/tex]

How many grams of water [tex]$\text{H}_2\text{O}$[/tex] will be formed when 12 grams of oxygen are reacted?

A. 8.10 grams
B. 38.2 grams
C. 87.4 grams
D. 98.3 grams

To solve this problem, we need to determine how many grams of water ([tex]$ H_2O $[/tex]) will be formed when 12 grams of oxygen ([tex]$ O_2 $[/tex]) are used in the given chemical reaction:

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

Here's a step-by-step solution:

1. Identify Molar Masses:
- The molar mass of [tex]$ O_2 $[/tex] is 32 grams/mol (since each oxygen atom has a molar mass of 16 grams/mol and [tex]$ O_2 $[/tex] has two oxygen atoms).
- The molar mass of [tex]$ H_2O $[/tex] is 18 grams/mol (16 grams/mol for oxygen and 2 grams/mol for two hydrogen atoms).

2. Calculate Moles of [tex]$ O_2 $[/tex]:
- Given 12 grams of [tex]$ O_2 $[/tex], calculate the number of moles of [tex]$ O_2 $[/tex]:
[tex]\[
\text{Moles of } O_2 = \frac{12 \text{ grams}}{32 \text{ grams/mol}} = 0.375 \text{ moles}
\][/tex]

3. Use the Reaction Stoichiometry:
- According to the balanced chemical equation, 5 moles of [tex]$ O_2 $[/tex] produce 6 moles of [tex]$ H_2O $[/tex].

4. Calculate Moles of [tex]$ H_2O $[/tex]:
- With 0.375 moles of [tex]$ O_2 $[/tex], the moles of [tex]$ H_2O $[/tex] formed can be calculated using the ratio from the balanced equation:
[tex]\[
\text{Moles of } H_2O = 0.375 \text{ moles } O_2 \times \frac{6 \text{ moles } H_2O}{5 \text{ moles } O_2} = 0.45 \text{ moles}
\][/tex]

5. Calculate Grams of [tex]$ H_2O $[/tex]:
- Finally, convert the moles of [tex]$ H_2O $[/tex] to grams:
[tex]\[
\text{Grams of } H_2O = 0.45 \text{ moles} \times 18 \text{ grams/mol} = 8.1 \text{ grams}
\][/tex]

Thus, 8.1 grams of water ([tex]$ H_2O $[/tex]) will be formed. Therefore, the correct answer is (a) 8.10 grams.