2Na3PO4(aq)+3 Ba(NO3)2(aq)---->Ba3(PO4)2(s)+6NaNO3(aq)

a solution containing 3.50g of Na3PO4 is mixed with a solution containing 6.40g of Ba(NO3)2 how many grams of product can be formed?
and if the above reaction resulted in 3.659 g of white powder, what is the percent yeild for the reaction?

This is a limiting reagent problem. You know that because amounts for BOTH reactants are given. I solve these, the long way by the way, by solving for moles product using the first reactant (ignore the second one), then the second reactant (ignore the first one). You will obtain two answers for the product and both can't be right, of course. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Here is a link that will show you exactly how to solve for the product. It also shows, at the end of the worksheet, how to determine percent yield.

http://www.jiskha.com/science/chemistry/stoichiometry.html
Post you work if you get stuck.

To find the grams of product formed, we need to determine the limiting reactant first. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be obtained.

Step 1: Calculate the number of moles for each reactant.
Given:
- Mass of Na3PO4 = 3.50g
- Mass of Ba(NO3)2 = 6.40g
- Molar mass of Na3PO4 = 163.94 g/mol
- Molar mass of Ba(NO3)2 = 261.34 g/mol

Number of moles of Na3PO4 = Mass / Molar mass
Number of moles of Na3PO4 = 3.50g / 163.94 g/mol = 0.0214 mol

Number of moles of Ba(NO3)2 = Mass / Molar mass
Number of moles of Ba(NO3)2 = 6.40g / 261.34 g/mol = 0.0245 mol

Step 2: Determine the ratio of moles of reactants based on the balanced equation.
From the balanced equation: 2Na3PO4 + 3Ba(NO3)2 → Ba3(PO4)2 + 6NaNO3
The ratio of moles is 2:3 for Na3PO4 to Ba(NO3)2.

Step 3: Identify the limiting reactant.
To find the limiting reactant, compare the ratio of moles of reactants.
For Na3PO4, we have 0.0214 mol and for Ba(NO3)2, we have 0.0245 mol.
Since the ratio is 2:3, we can calculate the equivalent moles for Na3PO4 using the ratio of 2/3:

Equivalent moles of Na3PO4 = (2/3) * 0.0245 mol = 0.0163 mol

Comparing the moles of Na3PO4 and Ba(NO3)2, we can see that Na3PO4 is the limiting reactant since it has fewer moles.

Step 4: Calculate the moles of the product formed.
From the balanced equation: 2Na3PO4 + 3Ba(NO3)2 → Ba3(PO4)2 + 6NaNO3
The mole ratio between Na3PO4 and Ba3(PO4)2 is 2:1.

Moles of Ba3(PO4)2 formed = (1/2) * 0.0163 mol = 0.00815 mol

Step 5: Calculate the mass of the product formed using the molar mass of Ba3(PO4)2:
Molar mass of Ba3(PO4)2 = 601.93 g/mol

Mass of Ba3(PO4)2 formed = Moles * Molar mass
Mass of Ba3(PO4)2 formed = 0.00815 mol * 601.93 g/mol = 4.909 g

Therefore, the mass of product formed is 4.909 grams.

To calculate the percent yield, we use the formula:
Percent Yield = (Actual Yield / Theoretical Yield) * 100

Given:
- Actual Yield = 3.659 g
- Theoretical Yield (calculated above) = 4.909 g

Percent Yield = (3.659 g / 4.909 g) * 100 = 74.60%

Therefore, the percent yield for the reaction is 74.60%.

To determine how many grams of product can be formed, we need to use stoichiometry.

1. Start by calculating the number of moles of Na3PO4 and Ba(NO3)2 using their molar masses:
moles of Na3PO4 = mass / molar mass = 3.50 g / (3(22.99 g/mol) + (31.00 g/mol) + 4(16.00 g/mol)) = 0.025 mol
moles of Ba(NO3)2 = mass / molar mass = 6.40 g / ((137.33 g/mol) + 2(14.01 g/mol) + 6(16.00 g/mol)) = 0.025 mol

2. Determine the limiting reactant. Since Na3PO4 and Ba(NO3)2 have a 1:3 molar ratio, and both have 0.025 mol, none of the reactants will be in excess. The limiting reactant will be Ba(NO3)2 because it gives the lower theoretical yield.

3. Next, find the molar ratio between Ba(NO3)2 and Ba3(PO4)2:
3 mol Ba(NO3)2 : 1 mol Ba3(PO4)2

4. Use the molar ratio to find the number of moles of Ba3(PO4)2 that can form:
moles of Ba3(PO4)2 = 0.025 mol Ba(NO3)2 × (1 mol Ba3(PO4)2 / 3 mol Ba(NO3)2) = 0.00833 mol Ba3(PO4)2

5. Calculate the mass of Ba3(PO4)2 using its molar mass:
mass of Ba3(PO4)2 = moles × molar mass = 0.00833 mol × ((3 (137.33 g/mol)) + 2(30.97 g/mol) + 4(16.00 g/mol)) = 1.64 g

Therefore, the mass of Ba3(PO4)2 that can be formed is 1.64 grams.

To calculate the percent yield, we need the actual yield and the theoretical yield.

The actual yield is given as 3.659 g.

The theoretical yield is the maximum possible yield based on stoichiometry, which is 1.64 g.

Percent yield = (actual yield / theoretical yield) × 100
Percent yield = (3.659 g / 1.64 g) × 100 = 223.66%

Therefore, the percent yield for the reaction is 223.66%.