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. In an experiment, 1.90 g of NH3 reacts with 4.96 g of O2. 4NH3(g) + 5O2(g) ⟶ 4NO(g) + 6H2O(g) (i) Which is a limiting reactant? Show your working. (ii) How many grams of excess reactant remain? Show your working. (iii) How many grams of NO is formed? Show your working

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Eduard22sly

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

Balanced equation

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

1. How to determine the limiting reactant

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.

2. How to determine the mass of the excess reactant remaining

  • 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

3. How to determine the mass of NO produced

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.

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