Equilibrium occurs when forward and reverse directions of a reversible reaction occur at the same rate so there is no overall change in the amounts of reactants and products.
Answer:
1) Maximun ammount of nitrogen gas: 
2) Limiting reagent: 
3) Ammount of excess reagent: 
Explanation:
<u>The reaction </u>

Moles of nitrogen monoxide
Molecular weight: 


Moles of hydrogen
Molecular weight: 


Mol rate of H2 and NO is 1:1 => hydrogen gas is in excess
1) <u>Maximun ammount of nitrogen gas</u> => when all NO reacted


2) <u>Limiting reagent</u>:
3) <u>Ammount of excess reagent</u>:


Sodium bicarbonate and acetic acid are not good substitute for sodium azide in airbags since the require more mass and produce less gas.
<h3>Which is the better chemical for an airbag?</h3>
The chemical equation for the production of nitrogen gas from sodium azide is given below:
1 mole or 66 go of sodium azide produces 3 moles or 67.2 L of nitrogen gas.
The equation for the production of carbon dioxide from sodium bicarbonate and acetic acid is given below:
- Na₂CO₃ + CH₃COOH → CH₃COONa + CO₂ + H₂O
1 mole, 106 g of Na₂CO₃ and 1 mole, 82 g of CH₃COOH are required to produce 1 mole or 22.4 L of CO₂.
The mass of sodium azide required is less than that of sodium bicarbonate and acetic acid required. Also, sodium azide produces a greater volume of gas. Therefore, sodium bicarbonate and acetic acid are not good substitute for sodium azide in airbags.
In conclusion, sodium azide is a better choice in airbags.
Learn more about airbags at: brainly.com/question/14954949
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Mercury-199 is composed of 80 protons, 119 neutrons, and 80 electrons.