Rachel adds together 450.9g of a 22.0% solution of potassium carbonate and 275 grams of a 5.57m solution of Barium chlorate. How many grams of precipitate does she produce?

To determine the number of grams of precipitate produced, we need to find the reactants that will form a precipitate when mixed. In this case, we are mixing a solution of potassium carbonate (K2CO3) and a solution of barium chlorate (Ba(ClO3)2).

To determine if a precipitate will form, we must check if the products of the reaction are both insoluble in water. In this case, if the products are barium carbonate (BaCO3) and potassium chloride (KCl), we need to verify their solubility.

Barium carbonate (BaCO3) is insoluble in water, which means it forms a precipitate. Potassium chloride (KCl), on the other hand, is soluble in water, indicating it will stay in solution.

Now, let's calculate the moles of each reactant to determine the limiting reactant, which will tell us which reactant is going to be fully consumed and thus determine the maximum amount of precipitate formed.

For the potassium carbonate solution:
Mass of potassium carbonate solution = 450.9 g
Concentration of potassium carbonate (K2CO3) = 22.0% solution

We can convert the mass of the solution to the mass of K2CO3:
Mass of K2CO3 = Mass of solution × (Concentration / 100)
= 450.9 g × (22.0 / 100)
= 450.9 g × 0.22
= 99.198 g

To convert grams to moles, we need to know the molar mass of K2CO3, which is calculated by adding the atomic masses of its components:
Molar mass of K2CO3 = 2 × atomic mass of K + atomic mass of C + 3 × atomic mass of O
= 2 × 39.1 g/mol + 12.0 g/mol + 3 × 16.0 g/mol
= 138.2 g/mol

Now, we can calculate the moles of K2CO3:
Moles of K2CO3 = Mass of K2CO3 / Molar mass of K2CO3
= 99.198 g / 138.2 g/mol
≈ 0.718 mol

For the barium chlorate solution:
Mass of barium chlorate solution = 275 g
Concentration of barium chlorate (Ba(ClO3)2) = 5.57 M solution

To convert moles to grams, we need to know the molar mass of Ba(ClO3)2:
Molar mass of Ba(ClO3)2 = atomic mass of Ba + 2 × (atomic mass of Cl) + 6 × (atomic mass of O)
= 137.3 g/mol + 2 × 35.5 g/mol + 6 × 16.0 g/mol
= 304.3 g/mol

To calculate the moles of Ba(ClO3)2:
Moles of Ba(ClO3)2 = Concentration of Ba(ClO3)2 × Volume (in L)
= 5.57 mol/L × (275 g / 304.3 g/mol)
≈ 5.02 mol

Now, using the balanced equation for the reaction between K2CO3 and Ba(ClO3)2:
K2CO3 + Ba(ClO3)2 -> BaCO3 + 2KCl

We can see that 1 mole of K2CO3 reacts with 1 mole of Ba(ClO3)2 to produce 1 mole of BaCO3. Therefore, the moles of BaCO3 formed would be equal to the moles of the limiting reactant.

In this case, since the moles of K2CO3 (0.718 mol) are less than the moles of Ba(ClO3)2 (5.02 mol), K2CO3 is the limiting reactant. This means all the K2CO3 will react, and Ba(ClO3)2 will be in excess.

Now, to calculate the mass of BaCO3 precipitate formed, we can use the molar mass of BaCO3:
Molar mass of BaCO3 = atomic mass of Ba + atomic mass of C + 3 × (atomic mass of O)
= 137.3 g/mol + 12.0 g/mol + 3 × 16.0 g/mol
= 197.3 g/mol

Moles of BaCO3 = Moles of limiting reactant (K2CO3) = 0.718 mol

Mass of BaCO3 = Moles of BaCO3 × Molar mass of BaCO3
= 0.718 mol × 197.3 g/mol
≈ 141.11 g

Therefore, Rachel produces approximately 141.11 grams of precipitate (BaCO3) by mixing the given solutions.