Let's understand the chemical reaction:
[tex]\[ 2 NO ( g )+ O _2( g ) \rightarrow 2 NO _2( g ) \][/tex]
From the balanced chemical equation, we can see the stoichiometric relationship between nitric oxide (NO) and nitrogen dioxide (NO₂):
- 2 moles of NO react with 1 mole of O₂ to produce 2 moles of NO₂.
This tells us that the molar ratio of NO to NO₂ is 1:1. In other words, for every mole of NO reacted, 1 mole of NO₂ is produced.
Now, let’s apply this ratio to the given problem where 2.00 moles of NO are reacted:
1. Identify the moles of NO reacted:
[tex]\[ 2.00 \text{ moles of NO} \][/tex]
2. Determine the molar ratio between NO and NO₂:
[tex]\[ \text{NO} : \text{NO₂} = 1 : 1 \][/tex]
3. Calculate the moles of NO₂ produced:
Since the ratio is 1:1, the moles of NO₂ produced will be the same as the moles of NO reacted. Thus:
[tex]\[ \text{Moles of NO₂ produced} = 2.00 \text{ moles of NO} \][/tex]
Therefore, if a sample of 2.00 moles of nitric oxide (NO) gas reacts with excess oxygen, it produces 2.00 moles of nitrogen dioxide (NO₂) gas.
So the answer is:
[tex]\[ \boxed{2.0 \text{ moles}} \][/tex]
