Question

You have x moles of Al2(SO4)3. Express each of the following quantities in terms of x.
a) The number of moles of aluminum ions
b) The number of moles of sulfate ions
c) The number of moles of oxygen atoms
d) The number of grams of Al2(SO4)3
e) The number of grams of oxygen

Answer 1

Answer:- (a) 2X mol, (b) 3X mol, (c) 12X mol, (d) 342.15X g and (e) 96X g

Solution:- (a) The dissociation equation for the given compound is written as:

$Al_2(SO_4)_3\rightarrow 2Al^+^3+3SO_4^-^2$

From this equation, there is 1:2 ratio between the aluminium sulfate and aluminum ion. So, the number of moles of Al ion in the X moles of the compound would be:

$Xmol Al_2(SO_4)_3(\frac{2mol Al ion}{1mol Al_2(SO_4)_3})$

= 2X moles of Al ion

So, there would be 2X moles of Al ions in the X moles of the compound.

(b) From the equation, there is 1:3 mol ratio between aluminum sulfate and sulfate ion. So, the number of moles of sulfate ion in the X moles of the compound would be:

$Xmol Al_2(SO_4)_3(\frac{3mol SO_4^-^2}{1mol Al_2(SO_4)_3})$

= $3Xmol SO_4^-^2$

So, there would be 3X moles of the sulfate ion in X moles of the compound.

(c) There are 12 moles of O atom in 1 mol of the compound. So, moles of O atom in X moles of compound would be:

$Xmol Al_2(SO_4)_3(\frac{12mol O atom}{1mol Al_2(SO_4)_3})$

= 12X moles of O atom

So, there would be 12X moles of O atom in X moles of the compound.

(d) Molar mass of aluminum sulfate is 342.15 gram per mol. It means the mass of 1 mol of the compound is 342.15 g. So, the mass of X moles of the compound would be:

$Xmol Al_2(SO_4)_3(\frac{342.15g}{1mol})$

= 342.15X g of $Al_2(SO_4)_3$

So, there are 342.15X g of aluminium sulfate in X moles of it.

(e) From part c, there are 12X moles of O in X moles of the compound and the atomic mass of O is 8. So, the grams of O would be:

$12X mol O(\frac{8g}{1mol})$

= 96X g of O

So, the mass of O in the X moles of the compound would be 96X g.