In order to purify ammonia to high purity, two basic methods are used. The older one consists in passing gaseous ammonia containing 80 ppm of impurities under atmospheric pressure through liquid ammonia with dissolved metallic sodium.
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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Answer:
35.47+10.1 is 45.57, I hope that helped nc I'm new to this app and yea.
Answer:
1.62 × 10²⁴ atoms are in 52.3 g of lithium hypochlorite.
Explanation:
To find the amount of atoms that are in 52.3 g of lithium hypochlorite, we must first find the amount of moles. We do this by dividing by the molar mass of lithium hypochlorite.
52.3 g ÷ 58.4 g/mol = 0.896 mol
Next we must find the amount of formula units, we do this be multiplying by Avagadro's number.
0.896 mol × 6.02 × 10²³ = 5.39 × 10²³ f.u.
Now to get the amount of atoms we can multiply the amount of formula units by the amout of atoms in one formula unit.
5.39 × 10²³ f.u. × 3 atom/f.u. = 1.62 × 10²⁴ atoms
1.62 × 10²⁴ atoms are in 52.3 g of lithium hypochlorite.
