I have a reaction here:

2Na3PO4.12H2O + 3BaCl.2H2O --> Ba3(PO4)2 + 6NaCl + 30H2O

and a question:

0.349g of BaCl2.2H2O and 0.624g of Na3PO4.12H2O are dissolved in 500mL of water to form a solution - how many grams of the excess reactant remain after the reaction has gone to completion?

I know that the reactant in excess is the sodium phosphate, and that the answer is 0.362g, but I can't figure out how to get there. Also, is the reaction wrong because the question says BaCl2 and the reaction I made has BaCl, but that's how it worked, so I'm just confused!

Actually, sorry, the amount LEFT OVER is .262g of Na3PO4 - .362 was the amount used.

To determine the grams of the excess reactant remaining after the reaction, we need to compare the stoichiometry of the reaction with the given amounts of reactants.

First, let's find the molar masses of the compounds involved:

Molar mass of BaCl2·2H2O: (137.33 g/mol) + 2[(1.01 g/mol) + 2(16.00 g/mol)] = 244.27 g/mol
Molar mass of Na3PO4·12H2O: 3(22.99 g/mol) + (31.00 g/mol) + 4(16.00 g/mol) + 12[(1.01 g/mol) + 2(16.00 g/mol)] = 380.12 g/mol

Next, we need to determine the number of moles of each reactant used:

Number of moles of BaCl2·2H2O = (mass of BaCl2·2H2O / molar mass of BaCl2·2H2O) = 0.349 g / 244.27 g/mol = 0.00143 mol
Number of moles of Na3PO4·12H2O = (mass of Na3PO4·12H2O / molar mass of Na3PO4·12H2O) = 0.624 g / 380.12 g/mol = 0.00164 mol

Looking at the balanced equation, the stoichiometric ratio between BaCl2·2H2O and Na3PO4·12H2O is 3:2. This means that for every 3 moles of BaCl2·2H2O, we need 2 moles of Na3PO4·12H2O. However, we have a 0.00143 mol of BaCl2·2H2O (which is less than what is needed) and 0.00164 mol of Na3PO4·12H2O (which is in excess).

To find the amount of excess reactant remaining, we need to determine how much of the Na3PO4·12H2O reacts. This can be done by multiplying the moles of Na3PO4·12H2O by the stoichiometric ratio from the balanced equation:

Moles of Na3PO4·12H2O that react = (0.00143 mol BaCl2·2H2O) × (2 mol Na3PO4·12H2O / 3 mol BaCl2·2H2O) = 0.000952 mol Na3PO4·12H2O

To find the remaining moles of Na3PO4·12H2O, subtract the moles that react from the initial moles:

Remaining moles of Na3PO4·12H2O = 0.00164 mol - 0.000952 mol = 0.000688 mol

Finally, we can find the mass of the excess reactant remaining:

Mass of excess reactant remaining = (remaining moles of Na3PO4·12H2O) × (molar mass of Na3PO4·12H2O)
= (0.000688 mol) × (380.12 g/mol) = 0.261 g ≈ 0.262 g

Therefore, the excess reactant, Na3PO4·12H2O, remaining after the reaction is approximately 0.262 grams.

Regarding your concern about the reaction, you are correct that the given reaction is not consistent with the reactant BaCl2.2H2O in the question. The reaction you provided (Ba3(PO4)2) is not balanced with the given reactants. However, since you mentioned that the answer is 0.362g, it seems that the balanced equation is not needed to determine the excess reactant remaining.

To determine how much of the excess reactant remains after the reaction, we need to calculate the amount of each reactant that is used up in the reaction and then compare it to the starting amount of each reactant.

First, let's determine the moles of each reactant:

Molar mass of BaCl2·2H2O:
(1 Ba) + (2 Cl) + (2 H2O) = 137.33 g/mol

Molar mass of Na3PO4·12H2O:
(3 Na) + (1 P) + (4 O) + (12 H2O) = 380.14 g/mol

Number of moles of BaCl2·2H2O:
Mass of BaCl2·2H2O = 0.349 g
Moles of BaCl2·2H2O = 0.349 g / 137.33 g/mol = 0.00254 mol

Number of moles of Na3PO4·12H2O:
Mass of Na3PO4·12H2O = 0.624 g
Moles of Na3PO4·12H2O = 0.624 g / 380.14 g/mol = 0.00164 mol

From the balanced equation, we can see that the molar ratio between BaCl2·2H2O and Na3PO4·12H2O is 3:2. This means that for every 3 moles of BaCl2·2H2O, we need 2 moles of Na3PO4·12H2O.

Comparing the moles of the two reactants, we can see that Na3PO4·12H2O is the limiting reactant (or the one that is completely consumed) because we have fewer moles of it compared to BaCl2·2H2O.

To find the amount of Na3PO4·12H2O that is fully consumed in the reaction, multiply the moles of Na3PO4·12H2O by its molar mass:

Mass of Na3PO4·12H2O fully consumed = 0.00164 mol * 380.14 g/mol = 0.623 g

Since the total mass of Na3PO4·12H2O used is 0.624 g, we can calculate the mass of Na3PO4·12H2O that remains unreacted:

Mass of Na3PO4·12H2O remaining = 0.624 g - 0.623 g = 0.001 g

However, it is important to note that the balanced equation you provided is incorrect. The correct balanced equation for the reaction is:

2Na3PO4·12H2O + 3BaCl2·2H2O -> Ba3(PO4)2 + 6NaCl + 30H2O

In this balanced equation, BaCl2·2H2O is used instead of BaCl. Make sure to double-check the chemical formulas and balance the equation again to ensure accuracy in future calculations.