3.5 grams of Sodium Phosphate is mixed with 6.4 grams of Barium Nitrate. How many grams of Barium Phosphate will be produced?
I work these limiting reagent problems (you know it is limiting reagent when BOTH reactants are given) by working TWO simple stoichiometry problems. By simple I mean they are not limiting reagent problems. Here is a worked example of a simple stoichiometry problem. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
By working two problems, one problem for each of the reactants, you will obtain two separate answers (different answers) for the amount of the product. The correct answer in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.
Post your work if you get stuck.
To find the number of grams of Barium Phosphate produced, we need to determine the limiting reactant in the reaction. The limiting reactant is the one that is completely consumed and determines the amount of product formed.
First, we'll calculate the number of moles for both Sodium Phosphate and Barium Nitrate using their respective molar masses.
The molar mass of Sodium Phosphate (Na3PO4) is:
Na: 22.99 g/mol x 3 = 68.97 g/mol
P: 30.97 g/mol
O: 16.00 g/mol x 4 = 64.00 g/mol
Total molar mass: 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol
To find the number of moles of Sodium Phosphate:
3.5 g / 163.94 g/mol = 0.021 moles
The molar mass of Barium Nitrate (Ba(NO3)2) is:
Ba: 137.33 g/mol
N: 14.01 g/mol x 2 = 28.02 g/mol
O: 16.00 g/mol x 6 = 96.00 g/mol
Total molar mass: 137.33 g/mol + 28.02 g/mol + 96.00 g/mol = 261.35 g/mol
To find the number of moles of Barium Nitrate:
6.4 g / 261.35 g/mol = 0.024 moles
Next, we'll examine the balanced chemical equation to determine the stoichiometry between the reactants and the product.
The balanced chemical equation for the reaction between Sodium Phosphate (Na3PO4) and Barium Nitrate (Ba(NO3)2) is:
3 Na3PO4 + 2 Ba(NO3)2 → 6 NaNO3 + Ba3(PO4)2
From the balanced equation, we can see that:
- 3 moles of Sodium Phosphate react with 2 moles of Barium Nitrate to produce 1 mole of Barium Phosphate.
Comparing the moles of reactants, we have:
Sodium Phosphate: 0.021 moles
Barium Nitrate: 0.024 moles
Since there is a 3:2 mole ratio between Sodium Phosphate and Barium Nitrate, we divide the moles of each by their respective coefficients to determine the limiting reactant.
For Sodium Phosphate:
0.021 moles / 3 = 0.007 moles (per 1 mole of Barium Phosphate)
For Barium Nitrate:
0.024 moles / 2 = 0.012 moles (per 1 mole of Barium Phosphate)
Since Barium Nitrate has a higher mole ratio per mole of Barium Phosphate, it is the limiting reactant.
Now, to find the number of grams of Barium Phosphate produced, we'll use the molar mass of Barium Phosphate (Ba3(PO4)2):
Ba: 137.33 g/mol x 3 = 411.99 g/mol
P: 30.97 g/mol x 2 = 61.94 g/mol
O: 16.00 g/mol x 8 = 128.00 g/mol
Total molar mass: 411.99 g/mol + 61.94 g/mol + 128.00 g/mol = 601.93 g/mol
Since the mole ratio between Barium Nitrate and Barium Phosphate is 1:1, the number of moles of Barium Nitrate (0.012 moles) is equal to the number of moles of Barium Phosphate produced.
Finally, to find the mass of Barium Phosphate:
Mass = Moles × Molar Mass
Mass = 0.012 moles × 601.93 g/mol
Mass = 7.22 grams
Therefore, 7.22 grams of Barium Phosphate will be produced.