Question

720g sol de CuSO4 , de c=40% reactioneaza cu 800g de NaOH, de c=20%.
Se cere:
a)masa de substanta introdusa in exces
b)masa bazei insolubile depuse
c)concentratia finala dupa indepartarea precipitatului

Answer 1

CuSO4 + 2 NaOH => Na2SO4 + Cu(OH)2 precipitat
a) md = 720*40/100 = 288 g CuSO4
md = 800*20/100= 80 g NaOH
160 g CuSO4..........2*40 g NaOH
x g CuSO4...............80 g NaOH
x = 160 g CuSO4
Din cele 288 g sulfat reactioneaza doar 160
m exces = 288 - 160 = 128 g CuSO4
b) 160 g CuSO4........98 g Cu(OH)2
160 g CuSO4.....................x
x = 98 g Cu(OH)2
c) c final = md final/ms final * 100
md final = m exces = 128 g
ms final = ms 1 - m reactionat = 720 - 128 = 592 g
c final = 128/592 * 100 = 21.62%