The mass in grams of NH₃ produced from the reaction is 3.4 g
<h3>Balanced equation</h3>
We'll begin by writing the balanced equation for the reaction. This illustrated below:
N₂ + 3H₂ -> 2NH₃
From the balanced equation above,
1 dm³ of N₂ reacted to produced 2 dm³ NH₃
<h3>How to determine the volume of NH₃ produced</h3>
From the balanced equation above,
1 dm³ of N₂ reacted to produced 2 dm³ NH₃
Therefore,
2.24 dm³ of N₂ will react to produce = 2.24 × 2 = 4.48 dm³ of NH₃
<h3>How to determine the mass of NH₃ produced</h3>
We'll begin by obtained the mole of 4.48 dm³ of NH₃. Details below:
22.4 dm³ = 1 mole NH₃
Therefore,
4.48 dm³ = 4.48 / 22.4
4.48 dm³ = 0.2 mole of NH₃
Finally, we shall determine the mass of NH₃ as follow:
- Molar mass of NH₃ = 17 g/mol
- Mole of NH₃ = 0.2 mole
- Mass of NH₃ =?
Mass = mole × molar mass
Mass of NH₃ = 0.2 × 17
Mass of NH₃ = 3.4 g
Learn more about stoichiometry:
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You have to start listing from the bottom :
3. Secondary Consumers
2. Primary Consumers
1. Producer
Answer:
Percent yield = 79.79 %
Explanation:
Given data:
Mass of Al₂O₃ = 821 g
Mass of Al = 349 g
Percent yield = ?
Solution:
Chemical equation:
2Al₂O₃ + 3C → 4Al + 3CO₂
Number of moles of Al₂O₃:
Number of moles = mass/molar mass
Number of moles = 821 g/ 101.96 g/mol
Number of moles = 8.1 mol
Now we will compare the moles of Al with Al₂O₃.
Al₂O₃ : Al
2 : 4
8.1 : 4/2×8.1 = 16.2 mol
Mass of Al:
Mass = number of moles × molar mass
Mass = 16.2 mol × 27 g/mol
Mass = 437.4 g
Percent yield:
Percent yield = (actual yield / theoretical yield)× 100
Percent yield = (349 g/ 437.4 g) × 100
Percent yield = 79.79 %
