Question

Intr-un vas in care se gasesc 200g solutie de H2SO4, de concentratie 80% se introduc 32g Cu (cupru) pur. Sa se calculeze concentratia procentuala finala.

PLS AM NEVOIE MAINE PLS

Answer 1

Cu + 2 H2SO4 => CuSO4 + SO2 + 2 H2O
c = md/ms * 100
md = ms*c/100 = 200*80/100 = 160 g H2SO4
64 g Cu............98 g H2SO4
32 g Cu..............x g H2SO4
x = 49 g H2SO4
m acid exces = 160 - 49 = 111 g acid
64 g Cu............2*18 g H2O
32 g Cu..............x g H2O
x = 18 g H2O
c final = md final/ms final * 100
md final = m exces acid = 111 g
ms final = ms1 + m exces acid + m apa formata
=> ms final = 200 + 111 + 18 = 329 g solutie
c = 111/329*100 = 33.738%