Show the calculation for the limiting reagent when 0.853g of sodium sulfate (95% pure) react with 1.707g of barium hydroxide octahydrate (87% pure) to produce a precipitate of barium sulfate.

I've determined the equation is Na2SO4+Ba(OH)2*8H2O=BaSO4+2NaOH

However, that is unbalanced and I'm not sure how to balance the equation with the hydrated compound.

What would the first steps be to balance each side?

You have it balanced except for the 8H2O. Just add 8H2O on the end.

Na2SO4+Ba(OH)2*8H2O=BaSO4+2NaOH + 8H2O
g Na2SO4 = 0.853 x 0.95 = ?
g Ba(OH)2*8H2O = 1.707 x 0.87 = ?

Convert to mols.
mols Na2SO4 = grams/molar mass = ?
mols Ba(OH)2*8H2O = grams/mola mass = ?

Using the coefficients in the balanced equation, convert mols Na2SO4 to mols either BaSO4 or NaOH.
Do the same and convert mols Ba(OH)2*8H2O to mols BaSO4 or NaOH (but use the same product for both calculations).It is likely that the two values will not agree; the correct value in limiting reagent problem is ALWAYS the smaller value and the reagent responsible for that is the limiting reagent.

To balance the equation, start by identifying the atoms that are unbalanced in the initial equation. In this case, we can see that there are 2 sodium atoms on the left side but only 1 sodium atom on the right side. Additionally, there is only 1 sulfate ion (SO4) on the left side, but it becomes 1 barium sulfate (BaSO4) on the right side.

To balance the sodium atoms, you need to add a coefficient of 2 in front of NaOH on the right side to match the 2 sodium atoms on the left side.

Na2SO4 + Ba(OH)2·8H2O → BaSO4 + 2NaOH

Regarding the sulfate ions, since you have 1 sodium sulfate (Na2SO4) on the left side, you need to have a coefficient of 1 in front of BaSO4 on the right side to balance the sulfate ions.

Na2SO4 + Ba(OH)2·8H2O → BaSO4 + 2NaOH

Now, we have a balanced equation.

Once you have the balanced equation, you can proceed to calculate the moles of each reactant and determine the limiting reagent.

To calculate the number of moles, you can use the formula:

moles = mass (g) / molar mass (g/mol)

For sodium sulfate (Na2SO4):

mass of Na2SO4 = 0.853 g
purity of Na2SO4 = 95% = 0.95
molar mass of Na2SO4 = 142.04 g/mol

moles of Na2SO4 = (mass of Na2SO4 * purity of Na2SO4) / molar mass of Na2SO4

For barium hydroxide octahydrate (Ba(OH)2·8H2O):

mass of Ba(OH)2·8H2O = 1.707 g
purity of Ba(OH)2·8H2O = 87% = 0.87
molar mass of Ba(OH)2·8H2O = 315.47 g/mol

moles of Ba(OH)2·8H2O = (mass of Ba(OH)2·8H2O * purity of Ba(OH)2·8H2O) / molar mass of Ba(OH)2·8H2O

With the moles calculated for each compound, you can then determine the limiting reagent by comparing the stoichiometric ratio of the reactants. The limiting reagent is the one that produces the least amount of product based on the stoichiometry of the balanced equation.