To determine the mass of nitrogen monoxide (NO) produced when 2 moles of ammonia (NH3) react with excess oxygen (O2), we should follow a systematic approach involving stoichiometry and the calculation of molar masses.
### Step 1: Write the balanced chemical equation
The balanced chemical equation for the reaction is:
[tex]\[ 4 \text{NH}_3(g) + 5 \text{O}_2(g) \rightarrow 6 \text{H}_2\text{O}(g) + 4 \text{NO}(g) \][/tex]
### Step 2: Identify the molar ratio
From the balanced equation, we observe that 4 moles of NH3 react to produce 4 moles of NO:
[tex]\[ 4 \text{NH}_3 \rightarrow 4 \text{NO} \][/tex]
### Step 3: Determine the moles of NO produced
Given that 2 moles of NH3 are reacting:
[tex]\[ \frac{4 \text{ moles } \text{NO}}{4 \text{ moles } \text{NH}_3} = \frac{x \text{ moles } \text{NO}}{2 \text{ moles } \text{NH}_3} \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = 2 \text{ moles NO} \][/tex]
### Step 4: Calculate the molar masses
- Molar mass of NH3:
[tex]\[ \text{M}_\text{NH}_3 = 14 \text{ (N)} + 3 \times 1 \text{ (H)} = 17 \text{ g/mol} \][/tex]
- Molar mass of NO:
[tex]\[ \text{M}_{\text{NO}} = 14 \text{ (N)} + 16 \text{ (O)} = 30 \text{ g/mol} \][/tex]
### Step 5: Calculate the mass of NO produced
Since we have 2 moles of NO produced:
[tex]\[ \text{mass of NO} = \text{moles of NO} \times \text{molar mass of NO} \][/tex]
[tex]\[ = 2 \text{ moles} \times 30 \text{ g/mol} \][/tex]
[tex]\[ = 60 \text{ g} \][/tex]
### Conclusion
The mass of NO produced when 2 moles of NH3 react with excess O2 is 60 grams.
