Determine the maximum g of Ba3(PO4)2 that can be formed from your mass of Na3PO4 . 12H2O
Here is a step by step procedure. You didn't provide all of the data in the post.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the maximum amount of Ba3(PO4)2 that can be formed from a given mass of Na3PO4 · 12H2O, we need to follow a few steps:
1. Write the balanced equation: Ba(NO3)2 + Na3PO4 → Ba3(PO4)2 + 6NaNO3. This equation shows the reaction between Ba(NO3)2 (barium nitrate) and Na3PO4 (sodium phosphate) to form Ba3(PO4)2 (barium phosphate) and NaNO3 (sodium nitrate).
2. Calculate the molar mass of Na3PO4 · 12H2O:
- Na (sodium) has a molar mass of 22.99 g/mol.
- P (phosphorus) has a molar mass of 30.97 g/mol.
- O (oxygen) has a molar mass of 16.00 g/mol.
- H (hydrogen) has a molar mass of 1.01 g/mol.
- Combining all these values, we get:
Molar mass of Na3PO4 · 12H2O = (22.99 g/mol × 3) + 30.97 g/mol + (16.00 g/mol × 4) + [(1.01 g/mol × 2) × 12] = 380.14 g/mol
3. Determine the number of moles of Na3PO4 · 12H2O:
- Use the given mass of Na3PO4 · 12H2O to calculate the number of moles using the formula:
Number of moles = Mass ÷ Molar mass
Assuming you have a specific mass, substitute the value for "Mass" in the formula.
4. Use the stoichiometric coefficients from the balanced equation to find the moles of Ba3(PO4)2:
The balanced equation shows that 1 mole of Na3PO4 reacts with 1 mole of Ba3(PO4)2. Therefore, the number of moles of Ba3(PO4)2 formed will be the same as the number of moles of Na3PO4 · 12H2O.
5. Calculate the mass of Ba3(PO4)2:
- Using the molar mass of Ba3(PO4)2, multiply the number of moles of Ba3(PO4)2 by its molar mass to obtain the mass of Ba3(PO4)2.
By following these steps, you'll be able to determine the maximum mass of Ba3(PO4)2 that can be formed from a given mass of Na3PO4 · 12H2O.