Traditional gravimetric analysis can be used to determine the composition of metals, such
as magnesium (Mg) and aluminium (Al), in some metal alloys. 8-hydroxyquinoline
(C9H7NO) is normally used as the precipitation reagent. In an analysis, a 1.85 g sample of
an alloy that contains magnesium (Mg), aluminium (Al), and other metals was dissolved
in acid and treated quantitatively with 21.788 g of 8-hydroxyquinoline (C9H7NO). A
mixed precipitate of Al(C9H6NO)3 and Mg(C9H6NO)2 is resulted, which is further ignited
to give a final mixture of Al2O3(s) and MgO(s) that weighs 3.006 g.
(a) Let x and y be the mass of Mg and Al in the original alloy sample respectively.
Manipulate two equations of the total mass of precipitate isolated and the amount of
8-hydroxylquinoline in the mixed precipitate in terms of x and y.
Chemistry?
2017-10-06 10:12 pm
回答 (1)
2017-10-09 12:36 am
First, list out the molar mass (unit: g/mol) of the cheimcals:
Mg = 24.3
Al = 27
8-hydroxyquinoline (C9H7NO) = 145.0
complex between Mg and 8-hydroxyquinoline (i.e. Mg(C9H7NO)2 ) = 24.3 + 145.0 * 2 = 314.3
complex between Al and 8-hydroxyquinoline (i.e. Al(C9H7NO)3 ) = 27.0 + 145.0 * 3 = 462.0
MgO = 40.3
Al2O3 = 102
Now, figure out the no. of mole of Mg & Al present:
no. of mole of Mg = mass of Mg / molar mass of Mg
= x / 24.3 (mole)
similarly, no. of mole of Al = y / 27.0
From the chemical formulae, we know that 1 mole of Mg-complex contains 2 moles of Mg,
i.e. mole ratio fo Mg : Mg-complex = 2 : 1
so, for (x / 24.3) mole of Mg, there is (x/24.3) / 2 mole of the Mg-complex (i.e. x/48.6) ;
similarly, mole ratio of Al : Al-complex = 3 : 1
for (y / 27.0) mole of Al, there is (y/27.0) / 3 mole of the Mg-complex (i.e. y/81.0).
Then, mass = no. of mole * molar mass.
So, mass of Mg-complex
= no. of mole of Mg-complex * molar mass of Mg-complex
= (x/48.6) * 314.3 = (314.3/48.6) * x
Similarly, mass of Al-complex
= (462.0/81.0) * y
And from the paragraph, we know the mass of both complexes is 21.788 g
Thus, (314.3/48.6) * x + (462.0/81.0) * y = 21.788
Now, apply the calculation to the oxides of the metals:
- no. of mole of MgO = no. of mole of Mg = x / 24.3
- no. of mole of Al2O3 = no. of mole of Al / 2 = y / 54.0
- mass of MgO = (40.3/24.3) * x
- mass of Al2O3 =(102.0/54.0) * y
- (40.3/24.3) * x + (102.0/54.0) * y = 3.006
Therefore, you have got a pair of simultaneous equations:
(314.3/48.6) * x + (462.0/81.0) * y = 21.788
(40.3/24.3) * x + (102.0/54.0) * y = 3.006
Solve them to obtain x & y.
Mg = 24.3
Al = 27
8-hydroxyquinoline (C9H7NO) = 145.0
complex between Mg and 8-hydroxyquinoline (i.e. Mg(C9H7NO)2 ) = 24.3 + 145.0 * 2 = 314.3
complex between Al and 8-hydroxyquinoline (i.e. Al(C9H7NO)3 ) = 27.0 + 145.0 * 3 = 462.0
MgO = 40.3
Al2O3 = 102
Now, figure out the no. of mole of Mg & Al present:
no. of mole of Mg = mass of Mg / molar mass of Mg
= x / 24.3 (mole)
similarly, no. of mole of Al = y / 27.0
From the chemical formulae, we know that 1 mole of Mg-complex contains 2 moles of Mg,
i.e. mole ratio fo Mg : Mg-complex = 2 : 1
so, for (x / 24.3) mole of Mg, there is (x/24.3) / 2 mole of the Mg-complex (i.e. x/48.6) ;
similarly, mole ratio of Al : Al-complex = 3 : 1
for (y / 27.0) mole of Al, there is (y/27.0) / 3 mole of the Mg-complex (i.e. y/81.0).
Then, mass = no. of mole * molar mass.
So, mass of Mg-complex
= no. of mole of Mg-complex * molar mass of Mg-complex
= (x/48.6) * 314.3 = (314.3/48.6) * x
Similarly, mass of Al-complex
= (462.0/81.0) * y
And from the paragraph, we know the mass of both complexes is 21.788 g
Thus, (314.3/48.6) * x + (462.0/81.0) * y = 21.788
Now, apply the calculation to the oxides of the metals:
- no. of mole of MgO = no. of mole of Mg = x / 24.3
- no. of mole of Al2O3 = no. of mole of Al / 2 = y / 54.0
- mass of MgO = (40.3/24.3) * x
- mass of Al2O3 =(102.0/54.0) * y
- (40.3/24.3) * x + (102.0/54.0) * y = 3.006
Therefore, you have got a pair of simultaneous equations:
(314.3/48.6) * x + (462.0/81.0) * y = 21.788
(40.3/24.3) * x + (102.0/54.0) * y = 3.006
Solve them to obtain x & y.
收錄日期: 2021-04-24 00:44:23
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171006141249AAitzNl