S+O₂→SO₂
2SO₂+O₂→2SO₃
SO₃+H₂O→H₂SO₄ (gazul SO₃ este barbotat in apa si se obtine ,,sangele industirei'' - vitriolul)
notam 90 kg deasupra H₂O si masa moleculara a apei sub:
M(H₂O) = 2*A(H) + A(O) = 2*1 + 1*16 = 2+16 = 18
sub H₂SO₄ notam masa moleculara a sa, iar deasupra punem ,,x'':
M(H₂SO₄)=2*A(H)+1*A(S)+4*A(O)= 2*1+1*32+4*16=2+32+64=98
scoatem proportia :
[tex]\frac{90kg}{18kg} = \frac{x}{98kg}[/tex] ⇒ [tex]x=\frac{98*90kg}{18} = 490 kg H2SO4[/tex]
CaO+H₂O→Ca(OH)₂ (CaO NU se dizolva in H₂O, dar reactioneaza)
in cazul apei, procedam la fel cum am facut anterior, dar notam ,,y'' deasupra Ca(OH)₂ si masa moleculara sub:
M(Ca(OH)₂) = 1*A(Ca)+2*A(O)+2*A(H) = 40 + 2*16+2*1=40+32+2=74
scoatem proportia:
[tex]\frac{90kg}{18kg} = \frac{y}{74kg}[/tex]⇒[tex]y=\frac{74*90kg}{18} = 370kgCu(OH)2[/tex]
