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14,8 grams of a mixture containing Fe and Cu is dissolved in 126 grams of 48% HNO3 solution to form solution X ( no amonium salts). X reacts with 400 ml of 1M NaOH and 0,5M KOH solution to form precipitate Y and solution Z. Heat Y in the air to unchanged mass to form 20 grams of a mixture containing Fe2O3 and CuO. Vaporate Z to form solid mixture T. Heat T until they reached the unchanged mass of 42,86 grams. The percent concentration of Fe(NO3)3 in X has the closest value to:
A. 8,2
B. 7,9
C. 7.6
D. 6,9
please help me with this exercise!

 

[Isla Chemistry]

14,8 grams of a mixture containing Fe and Cu is dissolved in 126 grams of 48% HNO3 solution to form solution X ( no amonium salts). X reacts with 400 ml of 1M NaOH and 0,5M KOH solution to form precipitate Y and solution Z. Heat Y in the air to unchanged mass to form 20 grams of a mixture containing Fe2O3 and CuO. Vaporate Z to form solid mixture T. Heat T until they reached the unchanged mass of 42,86 grams. The percent concentration of Fe(NO3)3 in X has the closest value to:
A. 8,2
B. 7,9
C. 7.6
D. 6,9
please help me with this exercise!

Hypothetically , the mass of the Fe,Cu mixture is 14,8 (grams) and the mass of the Fe2O3,CuO mixture is 20 (grams)
-> 56x + 64y = 14,8 ; 160x/2 + 80y = 20
-> x = 0,15 ; y = 0,1 ( suppose that nFe = x mole ; nCu = y mole )
nHNO3 = 0,96 mole
nNaOH = 0,4 mole , nKOH = 0,2 mole
Heat T until they reached the unchanged mass of 42,86 grams -> 0,4 NaNO2 + 0,2 KNO2 -> mass of those solids = 44,6 > 42,86 -> The base (OH-) remains
The solids contains : Na+ , K+ , NO2- (a mole) and OH- (b mole)
-> a + b = 0,6 and 23.0,4 + 0,2.39 + 46a + 17b = 42,86
-> a = 0,54 ; b = 0,06
Therefore , nNO3 - = 0,54 < 3nFe + 2nCu ( 0,55 ) -> The production after reaction contains Fe(2+) x mole , Fe(3+) y mole ( HNO3 'll fully reacted by the way ) nFe = x + y = 0,15
Electrical-preserved -> 2x + 3y + 0,1.2 = 0,54 (2)
(1)(2) -> x = 0,11 ; y = 0,04
nN release from the reducing agent = 0,96 - 0,54 = 0,42 mole
Inside the gas , nO = z mole
Electron-preserved -> 0,11.2 + 0,04.3 + 0,1.2 + 2z = 0,42.5
-> z = 0,78
mX = mFe,Cu + mHNO3(aq.) - mN - mO = 122,44 (grams)
-> %CFe(NO3)3 = 7,9% (B)
P/S : Lần sau hỏi dich ra tiếng Việt nhé