gravimetric?

Question

Antimony is alloyed with lead to increase the rigidity of components used in the construction of lead storage batteries.
2.351 g of a particular metallic alloy, compounded of only Pb and Sb, can be quantitatively converted into a 2.736-g mixture of the oxides PbO2 and Sb2O4.
What was the percentage (by mass) of antimony in the alloy?

Answer 1

Pb atomic wt = 207.19
Sb atomic wt = 121.75
oxygen = 16 (close enough)

PbO2 =239.19 g/mole
Sb2O4 = 307.5 g/mole or 153.75g for SbO2
x moles of PbO2 and y moles of SbO2
x moles Pb and y moles of Sb

2.736 = x(239.19g/mole) + y(153.75g/mole)
2.351 = x(207.19g/mole) + y(121.75g/mole)

now add -207.19 time equation 1 to equation 2 eliminate the x variable, then solve for y.

Answer 2

To the very same problem I gave solution on Wednesday:

1.- First, we have to stablish the chemical reaction that took place. I propose:

Pb + Sb + O2 -------> PbO2 + Sb2O4

2.- This is a equation we can balance:

Pb + 2Sb + 3O2 ------> PbO2 + Sb2O4

3.- We can call :

x = amount of Pb in alloy
y = amount of Sb in alloy
w = amount of lead oxide
z = amount of antimony oxide

4.- From the data above: x + y = total weight of the alloy and
w + z = total weight of oxide mixture

x + y = 2.778 g --------(Eq.1)
w + z = 3.234 g -------(Eq.2)

5.- On the other hand, quantitatively reacted the alloy to get the oxides. We need the molecular masses of all components, to know what is the mass fraction of Pb and Sb in final products:

M(Pb)=207.2 g/mol
M(Sb)=121.7 g/mol
M(PbO2) =239.2 g/mol
M(Sb2O4)=307.4 g/mol

mass fraction Pb in oxide = 207.2/239.2 = 0.865
mass fraction Sb in oxide = (2x121.7)/307.4 = 0.791

So, the weight of lead in mixture oxides would be:
0.865w = x ......(Eq.3)

and the weight of antimony in mixture oxides would be:
0.791z= y .......(Eq.4)

6) We have to solve the system:

x + y = 2.778 --------(Eq.1)
w + z = 3.234 --------(Eq.2)
0.865w = x ------------(Eq.3)
0.791z = y -----------(Eq.4)

giving:

x = 2.570 g of Pb in the alloy
y = 0.208 g of Sb in the alloy

7) Finally, the percentage of antimony in alloy is:

% antimony in alloy = (0.208 g / 2.778 g) x 100% = 7.48%

That´s it!

Just check numbers.

Good luck!

Answer 3

Let mass of Antimony = x

So, mass of Pb = 2.351-x

Mass of Sb2O4 = 307.518*x/121.760

Mass of PbO2 = 240*(2.351-x)/208

Now, 307.518*x/121.760 + 240*(2.351-x)/208 = 2.736

Solve for x............