If 0.125 L of a 0.100 M solution of Ba(OH02 are mixed with 0.200 L of a 0.0750 M solution of H3PO4, the following reaction takes place.
3Ba(OH)2 + 2H3PO4 ---> Ba3(PO4)2 + 6H2O
What mass of barium phosphate is formed?
I got 2.51 g Ba3(PO4)2, is that correct? If not could someone explain to me how to go about answering this problem? Please and thank you..
That looks ok to me. Very good.
To find the mass of barium phosphate formed in the reaction, we need to use the concept of stoichiometry and the given concentrations of the solutions. Here's how to solve the problem step-by-step:
Step 1: Calculate the number of moles of Ba(OH)2 and H3PO4 using the given concentrations and volumes.
- For Ba(OH)2:
Moles of Ba(OH)2 = concentration (mol/L) x volume (L)
= 0.100 M x 0.125 L
- For H3PO4:
Moles of H3PO4 = concentration (mol/L) x volume (L)
= 0.0750 M x 0.200 L
Step 2: Determine the limiting reagent.
Compare the moles of Ba(OH)2 and H3PO4 calculated in the previous step. The reactant that has fewer moles is the limiting reagent. In this case, we see that H3PO4 has fewer moles.
Step 3: Use the stoichiometry of the balanced equation to find the number of moles of Ba3(PO4)2 formed.
From the balanced equation, we know that 2 moles of H3PO4 react with 3 moles of Ba(OH)2 to produce 1 mole of Ba3(PO4)2.
Since H3PO4 is in excess (not limiting), we use the moles of Ba(OH)2 as the basis for the stoichiometric calculation:
Moles of Ba3(PO4)2 = (moles of Ba(OH)2) x (1 mole of Ba3(PO4)2 / 3 moles of Ba(OH)2)
Step 4: Calculate the mass of Ba3(PO4)2 using the molar mass of Ba3(PO4)2.
The molar mass of Ba3(PO4)2 can be found by adding up the atomic masses of its elements. Let's represent it by "Molar mass of Ba3(PO4)2" in the next calculation.
Mass of Ba3(PO4)2 = (moles of Ba3(PO4)2) x (Molar mass of Ba3(PO4)2)
Step 5: Perform the calculation.
Plug in the known values into the equations from steps 1 to 4 and calculate the mass of Ba3(PO4)2.
Now, let's do the actual calculation using the given values:
- Moles of Ba(OH)2 = 0.100 M x 0.125 L = 0.0125 moles
- Moles of H3PO4 = 0.0750 M x 0.200 L = 0.015 moles
- The limiting reagent is Ba(OH)2.
- Moles of Ba3(PO4)2 = 0.0125 moles x (1 mole of Ba3(PO4)2 / 3 moles of Ba(OH)2) = 0.004167 moles
- Molar mass of Ba3(PO4)2 = Add the atomic masses of Ba, P, and O (using a periodic table)
- Let's assume the molar mass of Ba3(PO4)2 is X g/mol (a placeholder).
- 3 moles of Ba (atomic mass = atomic number) x atomic mass of Ba = 3X g/mol
- 2 moles of P (atomic mass = atomic number) x atomic mass of P = 2X g/mol
- 8 moles of O (atomic mass = atomic number) x atomic mass of O = 8X g/mol
- The molar mass of Ba3(PO4)2 = (3X + 2X + 8X) g/mol = 13X g/mol (This is just a representation of the molar mass until we calculate it)
- Finally, mass of Ba3(PO4)2 = 0.004167 moles x 13X g/mol
To determine the specific value, we need the atomic masses of Ba, P, and O from the periodic table, and then calculate the sum of these atomic masses. By substituting these values into the equation, you can calculate the mass of Ba3(PO4)2 formed.