To determine how many grams of NO (nitric oxide) are formed from the catalytic oxidation of ammonia (NH₃) with excess oxygen (O₂), let's walk through the problem step-by-step:
### Step 1: Write and Balance the Chemical Equation
The unbalanced reaction is:
[tex]\[ \text{NH}_3 + \text{O}_2 \rightarrow \text{NO} + \text{H}_2 \text{O} \][/tex]
Balanced chemical equation:
[tex]\[ 4 \text{NH}_3 + 5 \text{O}_2 \rightarrow 4 \text{NO} + 6 \text{H}_2 \text{O} \][/tex]
### Step 2: Calculate the Molar Masses of NH₃ and NO
The molar masses of the involved compounds are:
- NH₃: [tex]\( 1 \times 14.01 + 3 \times 1.01 = 17.03 \, \text{g/mol} \)[/tex]
- NO: [tex]\( 1 \times 14.01 + 1 \times 16.00 = 30.01 \, \text{g/mol} \)[/tex]
### Step 3: Convert Mass of NH₃ to Moles
To find the number of moles of NH₃:
[tex]\[ \text{mass of NH}_3 = 824 \, \text{g} \][/tex]
[tex]\[ \text{molar mass of NH}_3 = 17.03 \, \text{g/mol} \][/tex]
[tex]\[ \text{moles of NH}_3 = \frac{824 \, \text{g}}{17.03 \, \text{g/mol}} \approx 48.39 \, \text{moles} \][/tex]
### Step 4: Use the Mole Ratio to Determine Moles of NO
From the balanced equation, the mole ratio of NH₃ to NO is 1:1. Therefore, the number of moles of NO formed is the same as the moles of NH₃ used:
[tex]\[ \text{moles of NO} = 48.39 \, \text{moles} \][/tex]
### Step 5: Convert Moles of NO to Mass
To find the mass of NO formed:
[tex]\[ \text{molar mass of NO} = 30.01 \, \text{g/mol} \][/tex]
[tex]\[ \text{mass of NO} = \text{moles of NO} \times \text{molar mass of NO} \][/tex]
[tex]\[ \text{mass of NO} = 48.39 \, \text{moles} \times 30.01 \, \text{g/mol} \approx 1452.04 \, \text{g} \][/tex]
### Conclusion:
The mass of NO formed is approximately 1452 grams.
Thus, the correct choice among the given options is:
[tex]\[ 1454 \, \text{g} \][/tex] (Round to nearest value provided in the options)
Therefore, 1454 grams of NO are formed in the reaction.
