Question

masa de azot cuprins in 9,2 kilomoli NO2 ; masa de clor continut in 5 moli AgCl; masa de H care se gaseste in 12 moli acid sulfuros. Am nevoie de ajutor urgent. Multimesc!! ​

Answer 1

Răspuns:

Explicație:

9,2 kmoli NO2

MNO2= 14 + 2.16=46-----> 46g/moli

9,2kmoli= 9200 moli NO2

m =9200moli .46g/moli=432200 g NO2

46g NO2---------14g azot

432200g----------------x

x=432200.14 :46=128800g azot=128,800 kg

5moli AgCl

m=5moli.133,3g/moli=667,5 g AgCl

MAgCl= 108 +35,5=133,5-----> 133,5g/moli

133,5 g-----------35,5 g clor

667,5g--------------y

y=667,5 . 35,5 : 133,5=177,5 g clor

12moli H2SO3

mH2SO3= 12moli.82g/moli=984 g H2SO3

MH2SO3=2 +32 +3.16=82---->82g/moli

82g H2SO3--------2gH

984g------------------z

z=984. 2 : 82=24 g hidrogen

verifica calculele!