Question

Va rog imi trebuie neaparat

Answer 1

MSO₄+M₂(SO₄)₃ = 3M+384 (amestec)

%Me sau %M = m M/M amestec*100 ⇔ 30,43 = 3M/3M+384*100 ⇔ 208,71M = 11685,12 ⇒ M = 56g/mol , Fe

a) Formulele substantelor : FeSO₄ si Fe₂(SO₄)₃

b) x - nr moli FeSO₄ , y - nr moli Fe₂(SO₄)₃

x/y = 3/2 ⇒ x = 1,5y

M FeSO₄ = 152g/mol , MFe₂(SO₄)₃ = 400g/mol

152x+400y = m ⇔ 228y+400y = m ⇒ m = 628y

%FeSO₄ = mFeSO₄/m*100 = 228/628*100 = 36,3 
%Fe₂(SO₄)₃ = 400/628*100 = 63,69 

c) Mcristalohidrat (FeSO₄×nH₂O) = 152+18n g/mol

152+18n cristalohidrat .............................18ng H₂O
973g........................................................441g

(152+18n)441 = 18n*973 ⇔ 67032 = 9576n ⇒ n = 7

F.M ⇒ FeSO₄×7H₂O