addieharris12
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N2 + 2H2 --> N2H4

Part 1:
How many grams of hydrazine, N2H4, can be made from 10.66 g of N2 and excess H2?

Part 2:
How many grams of hydrazine, N2H4, can be made from excess N2 and 7.78 g of H2?

Part 3:
How many grams of N2 are required to make 99.69 g of hydrazine? Assume that H2 is in excess.

Respuesta :

Eduard22sly

Answer:

1. 12.18g of N2H4.

2. 62.24g of N2H4

3. 87.23g of N2

Explanation:

N2 + 2H2 —> N2H4

Molar Mass of N2 = 2x14 =28g/mol

Molar Mass of N2H4 = (2x14) + (1x4) = 28 + 4 = 32g/mol

Molar Mass of H2 = 2x1 = 2g /mol

Mass conc of H2 from the balanced equation = 2 x 2 = 4g

1. From the balanced equation,

28g of N2 produced 32g N2H4.

Therefore, 10.66 g of N2 will produce = (10.66x32)/28 = 12.18g of N2H4.

2. From the balanced equation,

4g of H2 produced 32g of N2H4.

Therefore, 7.78 g of H2 will produce = (7.78x32)/4 = 62.24g of N2H4

3. From the balanced equation,

28g of N2 produced 32g N2H4.

Therefore, Xg of N2 will produce 99.69 g of N2H4 i.e

Xg of N2 = (28x99.69)/32 = 87.23g

CintiaSalazar

Taking into account the reaction stoichiometry:

  • 12.18 grams of hydrazine, Nâ‚‚Hâ‚„, can be made from 10.66 g of Nâ‚‚ and excess Hâ‚‚.
  • 62.24 grams of hydrazine, Nâ‚‚Hâ‚„, can be made from excess of Nâ‚‚ and 7.78 g of Hâ‚‚.
  • 87.23 grams of Nâ‚‚ are required to make 99.69 g of hydrazine.

Reaction stoichiometry

In first place, the balanced reaction is:

N₂ + 2 H₂ → N₂H₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Nâ‚‚: 1 mole
  • Hâ‚‚: 2 moles
  • Nâ‚‚Hâ‚„: 1 mole

The molar mass of the compounds is:

  • Nâ‚‚: 28 g/mole
  • Hâ‚‚: 2 g/mole
  • Nâ‚‚Hâ‚„: 32 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

N₂: 1 mole ×28 g/mole= 28 grams

H₂: 2 moles ×2 g/mole= 4 grams

N₂H₄: 1 mole ×32 g/mole= 32 grams

Part 1

The following rule of three can be applied: if by reaction stoichiometry 28 grams of Nâ‚‚ form 32 grams of Nâ‚‚Hâ‚„, 10.66 grams of Nâ‚‚ form how much mass of Nâ‚‚Hâ‚„?

[tex]mass of N_{2} H_{4} =\frac{10.66 grams of N_{2}x 32 grams of N_{2} H_{4} }{28grams of N_{2}}[/tex]

mass of Nâ‚‚Hâ‚„= 12.18 grams

12.18 grams of hydrazine, Nâ‚‚Hâ‚„, can be made from 10.66 g of Nâ‚‚ and excess Hâ‚‚.

Part 2

The following rule of three can be applied: if by reaction stoichiometry 4 grams of Hâ‚‚ form 32 grams of Nâ‚‚Hâ‚„, 7.78 grams of Hâ‚‚ form how much mass of Nâ‚‚Hâ‚„?

[tex]mass of N_{2} H_{4} =\frac{7.78 grams of H_{2}x 32 grams of N_{2} H_{4} }{4grams of H_{2}}[/tex]

mass of Nâ‚‚Hâ‚„= 62.24 grams

62.24 grams of hydrazine, Nâ‚‚Hâ‚„, can be made from excess of Nâ‚‚ and 7.78 g of Hâ‚‚.

Part 3

The following rule of three can be applied: If by reaction stoichiometry 32 grams of Nâ‚‚Hâ‚„ are formed from 28 grams of Nâ‚‚, 99.69 grams of Nâ‚‚Hâ‚„ are formed from how much mass of Nâ‚‚?

[tex]mass of N_{2} =\frac{99.69 grams of N_{2} H_{4} x 28 grams of N_{2}}{32 grams of N_{2} H_{4}}[/tex]

mass of Nâ‚‚= 87.23 grams

Finally, 87.23 grams of Nâ‚‚ are required to make 99.69 g of hydrazine.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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