DrBob222
answered
mols Na3PO4 = ? mols Na^+ x (1 mol Na3PO4/3 mols Na^+) = ?
Then g Na3PO4 = mols x molar mass = ? g
Explain Bot
answered
1. First, let's calculate the molar mass of Na3PO4:
- The atomic mass of sodium (Na) is 22.99 g/mol.
- The atomic mass of phosphorus (P) is 30.97 g/mol.
- The atomic mass of oxygen (O) is 16.00 g/mol.
Thus, the molar mass of Na3PO4 can be calculated as follows:
(3 * 22.99 g/mol) + 30.97 g/mol + (4 * 16.00 g/mol) = 163.94 g/mol.
2. Next, let's calculate the number of moles of Na+ ions needed to achieve a concentration of 0.600 M in 550 mL of solution:
The concentration of Na+ ions in the solution is given in moles per liter (M).
Therefore, we need to convert 550 mL to liters by dividing it by 1000:
550 mL ÷ 1000 = 0.55 L.
The concentration of Na+ ions is 0.600 M, which means there are 0.600 moles of Na+ ions in 1 liter of solution.
Multiplying the concentration by the volume will give us the number of moles of Na+ ions:
0.600 mol/L * 0.55 L = 0.33 moles.
3. Finally, let's determine the mass of Na3PO4 required to provide 0.33 moles of Na+ ions:
The stoichiometry of the chemical formula Na3PO4 tells us that each mole of Na3PO4 contains 3 moles of Na+ ions.
Therefore, the number of moles of Na3PO4 needed is:
0.33 moles Na+ ions ÷ 3 moles Na3PO4 = 0.11 moles Na3PO4.
To convert moles to grams, we use the molar mass of Na3PO4 calculated earlier:
0.11 moles Na3PO4 * 163.94 g/mol = 18.03 grams (rounded to two decimal places).
Therefore, approximately 18.03 grams of Na3PO4 will be needed to produce 550 mL of a solution with a concentration of Na+ ions of 0.600 M.
