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How many grams of Na3PO4 will be needed to produce 425 mL of a solution that has a concentration of Na plus ions of 0.900 M?
Wiki User
∙ 13y ago
Let us find moles first.
Molarity = moles of solute/Liters of solution ( 750 ml = 0.750 Liters )
0.375 M Na2SO4 = moles Na2SO4/0.750 Liters
= 0.28125 moles Na2SO4
===================
0.28125 moles Na2SO4 (142.05 grams/1 mole Na2SO4)
= 39.95 grams Na2SO4 needed
---------------------------------------you do significant figures!
Wiki User
∙ 11y ago
Wiki User
∙ 13y agoI must assume that this is a decomposition reaction, though you do not tell me!
Balanced equation.
Na3PO4 -> 3Na(+) + PO4(3-)
Get moles Na ions.
Molarity = moles of solute/Liters of solution (425 ml = 0.425 Liters )
0.900 M Na(+) = mole Na(+)/0.425 Liters
= 0.3825 mole Na(+)
Drive against Na3PO4
0.3825 moles Na(+) (1 mole Na3PO4/3 moles Na(+))(163.94 grams/1 mole Na3PO4)
= 20.9 grams sodium phosphate needed
Wiki User
∙ 7y agoNa3PO4 ==> 3Na^+ + PO4^3-
Note 1 mole Na3PO4 produces 3 moles Na^+
moles Na3PO4 needed = 0.900 mol/L x 0.525 L x 1 mol/3 mol Na = 0.1575 moles
mass Na3PO4 = 0.1575 moles Na3PO4 x 164 g/mol = 25.83 grams = 25.8 g (3 sig figs)
Wiki User
∙ 10y ago45.1 g/mol
Wiki User
∙ 13y ago40.2g
Wiki User
∙ 11y ago1.18g
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