Question

A 50.0 mL sample of a 0.200 M aqueous solution of Na3PO4 was added to 50.0 mL of a 0.100 M aqueous solution of BaCl2. The mixture was stirred and the precipitate was collected, dried carefully, and weighed. How many grams of precipitate should be obtained

Answer 1

Answer:

1.02 grams

Explanation:

The reaction between Na₃PO₄ and BaCl₂ is:

2Na₃PO₄(aq) + 3BaCl₂(aq) → Ba₃(PO₄)₂(s) + 6NaCl(aq)            

We need to find the number of moles of Na₃PO₄ and BaCl₂:

$n_{Na_{3}PO_{4}} = C*V = 0.200 M*0.050 L = 0.01 moles$

$n_{BaCl_{2}} = C*V = 0.100 M*0.050 L = 0.005 moles$

Now, we need to find the limiting reactant:

$n_{Na_{3}PO_{4}} = \frac{2}{3}*0.005 moles = 0.0033 moles$

We have that it is needed 0.0033 moles of Na₃PO₄ to react with BaCl₂ and we have 0.01 moles of Na₃PO₄, so the limiting reactant is BaCl₂.

The number of moles of Ba₃(PO₄)₂ is:

$n_{Ba_{3}(PO_{4})_{2}} = \frac{0.005}{3} = 0.0017 moles$

Finally, the mass of Ba₃(PO₄)₂ is:

$m = n_{Ba_{3}(PO_{4})_{2}}*M = 0.0017 moles*601.93 g/mol = 1.02 g$

Therefore, should be obtained 1.02 grams of the precipitate.

I hope it helps you!