Agravimetric experiment was performed to determine the number of moles of sulfate that were present in a solution of sodium sulfate. in this experiment, an excess of barium nitrate was added to the solution and a precipitate formed. the resulting mixture was filtered and the precipitate was dried. the dry mass of the precipitate was measured to be 8.642 g. write the balanced net ionic equation for the precipitate reaction and calculate the number of moles of sulfate in the sample.

the balanced net ionic equation for the precipitate reaction

Na2SO4  + Ba(NO3)2 > BaSO4  + Na2(NO3)2

the number of moles of sulfate in the sample
M = m /n

M molar mass
m mass
n number of moles
so n =m/M

n = 8.642/233.33=0.037mol

1) Net ionic equation

Here you do not have a net ionic equation.

The net ionic equation is the ionic equation removing the spectator ions.

Spectator ions are those that are the same (do not change its oxidation number) in both side of the equation.

In this reaction all the ionic compounds keep the same oxdidation number, then you should remove all the species from the ionic equation which will let you with nothing.

This is the ionic equation:

2Na(1+) + SO4(2-) + Ba(2+) + 2NO3(1-)  --->2 Na (1+) + 2NO3(1-) + Ba(2+) + SO4(2-)

As you see all the ions, Na(1+), SO4(2-) , Ba(2+) and NO3(1-) are the same in both sides, which will let you with none in the net ionic equation.

2) Number of moles of sulphate in the sample.

All the precipitate is Barium Sulfate, Ba(SO4), because it is the insoluble product. NaNO3 is very soluble.

The the mass 8.642g is all Ba(SO4).

To convert grams to # of moles, you use the molar mass (MM) and this relation:

# moles = grams / MM

The molar mass of Ba SO4 is: 137.3 g/mol+ 32 g/mol + 4*16 g/mol = 233.3 g/mol

# moles = 8.642 g / 233.3 g/mol = 0.037 mol    
 
Which is the number of moles in the product.

By the stoichiometry, the number of moles of the sample is the same of the product.

The scale factor of what ?