Question

Am si eu nevoie de ajutor cu problema 2 din imagine.
Dau coronita si multe puncte!
Va rog mult!!

Answer 1

a.

ms= 150g , c% = 36,5%

c% = mdx100/ms => md = msxc%/100

= 150x36,5/100 = 54,75 g HCl

M.HCl = 1+35,5 = 36,5 g/mol

M.NaOH = 40+16+1 = 40 g/mol

M.NaCl = 23+35,5 = 58,5 g/mol

M.H2O = 2x1+16 = 18 g/mol

md g     54,75 g  a g      b g

NaOH + HCl --> NaCl + H2O

40        36,5      58,5      18

=> md = 40x54,75/36,5 = 60 g NaOH

din c% = mdx100/ms = 60x100/200 = 30%

b.

m.apa.sol.finala = m.apa.sol.NaOH + m.apa.sol.HCl + m.apa.formata(b)

m.apa.sol.NaOH = ms-md = 200-60 = 140 g

m.apa.sol.HCl = ms-md = 150-54,75 = 95,25 g

m.apa.formata = b = 54,75x18/36,5 = 27 g

=> m.apa.sol.finala = 140+95,25+27 = 262,25 g

c.

a = 54,75x58,5/36,5 = 87,825 g NaCl formata

d.

c%finala = md(NaCl)x100/ms.finala

ms.final = m.sare(a) + m.apa.sol.finala

= 87,825 + 262,25 =

=> c%finala = 87,825x100/350,075 = 25% (rotunjit)