Question

4,6g Na se introduc in 1000 g H20.Determina
a)masa de apa reactionata
b)nr de moli de gaz degajat
c)concentratia solutiei finale

Answer 1

M,Na= 23g/mol

M,NaOH= 40g/mol

2Na    +    2H2O----> 2NaOH  + H2

22x23g....2x18g..........2x40g.........1mol

4,6g...........x.......................y.................z

x= 3,6gH2O-----> m,apa ramasa=1000-3,6=> 996,4g

y= 8gNaOH

Z=0,2molH2

c=mdx100/ms      md= 8gNaOH

                              ms= md+m,apa= 8+996.4=. 1004,4g

c= 8x100/1004,4=.....................calculeaza !!!!!