20g H2 cu p=70%
p= m,px100/mi---> m,p= pxm,i/100= 14g H2pur
n,mol= m/M= 14g/2g/mol= 7mol
30gCl2 cu p=80%------->m,pur= 24g Cl2pur
n,mol= 24g/71g/mol=0,34mol
1mol 1mol 2mol
din ecuatia: H2 + Cl2-----> 2HCl
0,34......0,34........0,68
H2 este in exces : 7mol-0,34mol=6,66mol
amestecul final contine excesul de H2-6,66mol si HCl-o,68mol