1. The limiting reactant in the reaction is NH₃
2. The mass of the excess reactant remaining is 0.49 g
3. The mass of NO produced from the reaction is 3.35 g
<h3>Balanced equation </h3>
4NH₃ + 5O₂ —> 4NO + 6H₂O
Molar mass of NH₃ = 14 + (3×1) = 17 g/mol
Mass of NH₃ from the balanced equation = 4 × 17 = 68 g
Molar mass of O₂ = 16 × 2 = 32 g/mol
Mass of O₂ from the balanced = 5 × 32 = 160 g
Molar mass of NO = 14 + 16 = 30 g/mol
Mass of NO from the balanced equation = 4 × 30 = 120 g
SUMMARY
From the balanced equation above,
68 g of NH₃ reacted with 160 g of O₂ to produce 120 g of NO
<h3>1. How to determine the limiting reactant </h3>
From the balanced equation above,
68 g of NH₃ reacted with 160 g of O₂
Therefore,
1.90 g of NH₃ will react with = (1.90 × 160) / 68 = 4.47 g of O₂
From the calculation made above, we can see that only 4.47 g out of 4.96 g of O₂ given is needed to react completely with 1.90 g of NH₃.
Therefore, NH₃ is the limiting reactant.
<h3>2. How to determine the mass of the excess reactant remaining </h3>
- Mass of excess reactant (O₂) given = 4.96 g
- Mass of excess reactant (O₂) that reacted = 4.47 g
- Mass of excess reactant (O₂) remaining =?
Mass of excess reactant (O₂) remaining = 4.96 – 4.47
Mass of excess reactant (O₂) remaining = 0.49 g
<h3>3. How to determine the mass of NO produced </h3>
In this case, the limiting reactant (NH₃) will be used.
From the balanced equation above,
68 g of NH₃ reacted to produce 120 g of NO
Therefore,
1.90 g of NH₃ will react to produce = (1.90 × 120) / 68 = 3.35 g of NO.
Thus, 3.35 g of NO were obtained from the reaction.
Learn more about stoichiometry:
brainly.com/question/14735801
