100.0 g of Potassium Phosphate reacts with 100.0 g of Barium Chloride. Which molecule is completely used up by the reaction? How many grams of the other molecule are left over? When I started with 100.0 g of K3PO4 I got 147.1g of BaCl2. When I started with 100.0g of BaCl2 I got 67.97g of K3PO4. How do i tell which i use up and how many of the other are left????????
I work these limiting reagent (LR) problems the long way.
2K3PO4 + 3BaCl2 ==> Ba3(PO4)2 + 6KCl
mols K3PO4 = grams/molar mass = 100/212.26 = about 0.47
mols BaCl2 = 100/208.23 = about 0.48
Now convert mols K3PO4 to mols Ba3(PO4)2 using the coefficients in the balanced equation.
0.47 x (1 mol Ba3(PO4)2/2 mol K3PO4) = 0.47 x 1/2 = about 0.24
Do the same for mols BaCl2 to mols Ba3(PO4)2.
0.48 x (1 mol Ba3(PO4)2/3 mol BaCl2) = about 0.16
Of course both of those numbers can't be right; the correct value in limiting reagent (LR)problems is ALWAYS the smaller value and the reagent producing that number is the LR. It will be used completely and the other reagent will have some remaining; i.e., there will be an excess of the other reagent.
How to calculate the grams excess reagent remaining? First, calculate how much is used.
0.16 mol BaCl2 x (2 mol K3PO4/2 mol BaCl2) = 0.16 x 2/3 = about 0.11.
You had 0.16 initially, you've used 0.11 so you have 0.16-0.11 = about 0.05 mols remaining. Then g = mols x molar mass.
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To use your numbers it works this way.
If you use all 100 g BaCl2 it will take 67.1 g K3PO4. You have that much so BaCl2 will be limiting and K3PO4 will be in excess. If you started with 100 g K3PO4, you would need 147.1 g BaCl2 and you don't have that much; therefore, BaCl2 is the limiting reagent and K3PO4 is in excess. Actually, you only need to use either of those calculations to know which is which because either way you start the answer is the same. Another time saver is that you don't need to go through for grams. You can convert 100g BaCl2 and 100g K3PO4 to mols and use the same reasoning as above with mols. Hope this helps.
To determine which molecule is completely used up in the reaction and how much of the other molecule is left over, we need to compare the amount of each molecule obtained from the reactions.
First, let's calculate the molar masses of Potassium Phosphate (K3PO4) and Barium Chloride (BaCl2):
- The molar mass of K3PO4 = 3(atomic mass of K) + atomic mass of P + 4(atomic mass of O)
= 3(39.1 g/mol) + 31.0 g/mol + 4(16.0 g/mol)
= 94.3 g/mol + 31.0 g/mol + 64.0 g/mol
= 189.3 g/mol
- The molar mass of BaCl2 = atomic mass of Ba + 2(atomic mass of Cl)
= 137.3 g/mol + 2(35.4 g/mol)
= 137.3 g/mol + 70.8 g/mol
= 208.1 g/mol
Now, let's use the information given to determine which molecule is completely used up and how many grams of the other molecule are left over:
1. From the first reaction, starting with 100.0 g of K3PO4, you obtained 147.1 g of BaCl2.
- To calculate the number of moles of K3PO4, divide the given mass by its molar mass:
Moles of K3PO4 = 100.0 g / 189.3 g/mol = 0.528 mol
- To calculate the number of moles of BaCl2, divide the obtained mass by its molar mass:
Moles of BaCl2 = 147.1 g / 208.1 g/mol = 0.707 mol
2. From the second reaction, starting with 100.0 g of BaCl2, you obtained 67.97 g of K3PO4.
- To calculate the number of moles of BaCl2, divide the given mass by its molar mass:
Moles of BaCl2 = 100.0 g / 208.1 g/mol = 0.480 mol
- To calculate the number of moles of K3PO4, divide the obtained mass by its molar mass:
Moles of K3PO4 = 67.97 g / 189.3 g/mol = 0.359 mol
By comparing the number of moles in each reaction, we can determine which molecule is completely used up:
- In the first reaction: Moles of K3PO4 = 0.528 mol, Moles of BaCl2 = 0.707 mol
- In the second reaction: Moles of BaCl2 = 0.480 mol, Moles of K3PO4 = 0.359 mol
From the calculations, it is clear that K3PO4 is completely used up in both reactions, and BaCl2 is left over.
To find out how much BaCl2 is left over, we can subtract the amount used in the reaction from the initial amount:
In the first reaction: BaCl2 used = 0.707 mol - 0.480 mol = 0.227 mol
Mass of BaCl2 used = Moles of BaCl2 used * molar mass of BaCl2
= 0.227 mol * 208.1 g/mol = 47.28 g
Therefore, if you started with 100.0 g of K3PO4, you completely used it up, and 47.28 g of BaCl2 is left over.
I hope this explanation helps you understand how to determine which molecule is used up and how much of the other molecule is left over in a reaction by comparing the amounts obtained from each reaction.