Osnovi hemijskog racunanja

Preko masenog udela izracunas masu nitritne (azotaste) kiseline u rastvoru-maseni udeo*masa rastvora.  Zatim stehiometrijski preracunas zapreminu izdvojenog NO, ako znas da pri normalnim uslovima 1 MOL GASA IMA ZAPREMINU 22,4 dm³

"A new scientific truth does not triumph by convincing opponents and making them see the light, but rather because its opponents eventually die, and a new generation grows up that is familiar with it" Max Planck

Sve sto si rekao i ja znam, nego ne znam kako da napisem. Ako bih mogao da mi pokazes, bio bih ti veoma zahvalan!

Unapred ti hvala!

Pa ne znam sta je onda problem. Nadjes masu HNO2: 0.2*250=50g
Imas vec gotovu reakciju:
2HNO2 ----> HNO3 + 2NO + H2O
...50g..............................x dm3
..94g................................44,8 dm3
x=23,83dm3

"A new scientific truth does not triumph by convincing opponents and making them see the light, but rather because its opponents eventually die, and a new generation grows up that is familiar with it" Max Planck

Kako moze da bude izjednacena reakcija:

2HNO2 ----> HNO3 + 2NO + H20.

??? Trebalo bi da bude :

3HNO2 ----> HNO3 + 2NO + H20

Sto bi promenilo rezultat alexa996:

50g : x dm3 =141 g : 44,8 dm3

x=15,89 dm3 NO

Nisam ni obracao paznju na reakciju, veliki previd.

"A new scientific truth does not triumph by convincing opponents and making them see the light, but rather because its opponents eventually die, and a new generation grows up that is familiar with it" Max Planck