Question

Peste 1,6g sulfat de cupru se adauga solutie de NaOH.Care este raportul molar dintre hidroxidul de sodiu consumat si hidroxidul de cupru rezultat?
Va roggg!!!

Answer 1

mCuSO4=1,6g

MCuSO4=ACu+AS+4AO=64+32+4*16=160g/mol

MNaOH=ANa+AO+AH=23+16+1=40g/mol

MCu(OH)2= ACu+2(AO+AH) = 64+2(16+1) = 98g/mol


80g..............160g.................................98g
2NaOH + CuSO4 => Na2SO4 + Cu(OH)2
x=0,8g...........1,6g...........................y=0,98g

n=m/M=0,8/40=0.02 moli NaOH

n=m/M=0,98/98=0,01 moli

nNaOH/nCu(OH)2 = 0,02/0,01 = 2/1