Question

Cati moli de NH3 se obtin din H dezlocuit in reactia Zn cu 300g solutie HCl de concentratie=7,3%

Answer 1

C=md/ms*100=>md=C*ms/100=7,3*300/100=21,9g HCl

n=m/M=21,9/36,5=0,6 moli HCl

Zn + 2HCl ---> ZnCl2 + H2

2 moli HCl........1 mol H2
0,6 moli HCl.....x=0,3 moli H2

N2 + 3H2 ---> 2NH3

3 moli H2.......2 moli NH3
0,3 moli H2....x=0,2 moli NH3