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A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4 will be needed to produce 725 mL of a solution that has a concentration of Na ions of 1.30 M

Respuesta :

Answer:

51.53 grams .

Explanation:

Na₃PO₄    ⇄  3Na⁺¹   +    PO₄⁻³ .

1 mole           3 mole

725 mL of 1.3 M Na⁺  ions

= .725 x 1.3 moles of Na⁺ ions

= .9425 moles

3 mole of Na⁺  is formed by 1 mole of Na₃PO₄

.9425 mole of Na⁺  is formed by .9425/3  mole of Na₃PO₄

Na₃PO₄ needed =  .9425/3  moles = .3142 moles

Molecular weight of Na₃PO₄ = 164

grams of Na₃PO₄ needed = .3142 x 164 = 51.53 grams .

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