If you begin with 100.0 mL of a 3.0 Molar solution of ammonium sulfate and 100mL of a 2.0 Molar solution of barium phosphate how many grams of Barium sulfate would be produced?

(NH4)2 + Ba3(PO4)2 ==> BaSO4 + (NH4)3PO4

Note: Ba3(PO4)2 is quite insoluble; however, it is more soluble than BaSO4 and I wrote the equation to reflect that.
Convert 100 mL of 3.0 M (NH4)2SO4 to moles. moles = M x L.
Do the same for the barium phosphate.

Using the coefficients in the balanced equation, convert mole ammonium sulfate to moles BaSO4. Do the same for moles barium phosphate to BaSO4. The smaller value of moles for BaSO4 produced will be the correct value to use.
g BaSO4 = moles BaSO4 x molar mass BaSO4.

Is this correct:

H3N+MgOH2=H5O+MgN

To determine the number of grams of barium sulfate produced, you first need to determine the limiting reactant between ammonium sulfate (NH4)2SO4 and barium phosphate (Ba3(PO4)2). The limiting reactant is the one that is completely consumed and determines the amount of product formed.

Let's start by calculating the moles of ammonium sulfate and barium phosphate:

Moles of ammonium sulfate = volume (L) x molarity (mol/L)
Moles of ammonium sulfate = 0.100 L x 3.0 mol/L = 0.300 mol

Moles of barium phosphate = volume (L) x molarity (mol/L)
Moles of barium phosphate = 0.100 L x 2.0 mol/L = 0.200 mol

Next, we need to determine the stoichiometric ratio between barium sulfate (BaSO4) and barium phosphate. From the balanced chemical equation:

3 Ba3(PO4)2 + 8 (NH4)2SO4 → 16 NH3 + 4 H3PO4 + 6 BaSO4

We can see that 3 moles of barium phosphate produce 6 moles of barium sulfate.

Now, compare the moles of barium sulfate that can be produced from both reactants. Since it takes 3 moles of barium phosphate to produce 6 moles of barium sulfate, we can determine that only 0.200 mol x (6 mol/3 mol) = 0.400 mol of barium sulfate can be produced.

Finally, calculate the mass of barium sulfate using its molar mass:

Mass of barium sulfate = moles of barium sulfate x molar mass of BaSO4
Mass of barium sulfate = 0.400 mol x 233.39 g/mol = 93.36 grams

Therefore, approximately 93.36 grams of barium sulfate would be produced in this reaction.