Respuesta :
Answer : The value of the scaling factor by which all reagents reduced must be 0.0711
Explanation :
First we have to calculate the molecular weight of alum.
The formula of alum is, [tex]KAl(SO_4)_2.12H_2O[/tex]
Molecular weight of Alum = [tex]39+27+2\times (32+16\times 4)+12\times (2\times1+16)=474g/mol[/tex]
Now we have to calculate the moles of alum.
[tex]\text{Moles of }PbS=\frac{\text{Mass of }PbS}{\text{Molar mass of }PbS}=\frac{15.0g}{474g/mole}=0.0316mole[/tex]
Now we have to calculate the moles of aluminium.
From the molecular formula of the alum we conclude that, 1 mole of alum constitutes 1 mole of aluminium.
So, 0.0316 moles of alum constitutes the number of moles of aluminium = 0.0316 moles
Now we have to calculate the mass of aluminum.
Molar mass of aluminium = 27 g/mole
[tex]\text{ Mass of }Al=\text{ Moles of }Al\times \text{ Molar mass of }Al[/tex]
[tex]\text{ Mass of }Al=(0.0316moles)\times (27g/mole)=0.853g[/tex]
Now we have to calculate the value of the scaling factor.
In the the original procedure, the amount of aluminium required = 12.0 g
The amount of aluminium required to produce 15 gram alum = 0.853 g
The scaling factor = [tex]\frac{0.853}{12}=0.0711[/tex]
Therefore, the value of the scaling factor by which all reagents reduced must be 0.0711
