godreigns20
contestada

In the following reaction, how many grams of ammonia (NH3) will produce
300 grams of N2?
4NH3 + 6NO → 5N2 + 6H2O
The molar mass of ammonia is 17.0337 grams and that of nitrogen is 28.02 grams.

394.794 grams

227.967 grams

145.899 grams

616.865 grams

Respuesta :

145.899 grams

Take the 300 grams of N2 you want and divide by its molar mass of 17.0337 to tell you that you want 10.70664 moles of N2. Since for every 5 moles of N2 produced, you need 4 moles of ammonia, divide by 5 and then multiply by 4 to get the number of moles of ammonia you need. This will be 10.70664 / 5 * 4 = 8.56531 moles. Finally multiply by the molar mass of ammonia and you'll know how many grams of ammonia you need. This is 8.56531 * 17.0337 = 145.8989 grams of ammonia which rounds off to 145.899 grams.


Ifecosos

Answer:

The correct option is the third option which is 145.899g

Explanation:

From the equation of reaction: 4NH3 + 6NO → 5N2 + 6H2O

First calculate the masses involved in the given equation of reaction.

Mass of ammonia NH3 = 4 x 17.0337= 68.1348

Mass of nitrogen N2 = 5 x 28. 02 = 140.1

Use the following analogy

68.1348g of ammonia produced  140.1g of N2

Yg of ammonia will be produced by 300g of N2 (Using Y as unknown mass)

Calculating for Y by cross-multiplying

Therefore 300g of N2 (Y) will be produced by 68.1348 x 300/ 140.1 = 145.899g

Otras preguntas