What mass of precipitate (in g) is formed when 45.5 mL of 0.300 M Na3PO4 react with 42.5mL of 0.200 M Cr(NO3)3 in the following

Question

3 years ago

7

chemical reaction?
Na3PO4(aq)+ Cr(NO3)3(aq) -> CrPO4(s) +3NaNO3(aq)

1 answer:

Marysya12 [62] · Answer 1 · 2022-09-04T11:06:59+03:00

The mass of the precipitate, CrPO₄ formed when 45.5 mL of 0.300 M Na₃PO₄ react with 42.5mL of 0.200 M Cr(NO3)₃ is 1.25 g

We'll begin by calculating the number of mole of Na₃PO₄ and Cr(NO3)₃ present in the solution.

Molarity of Na₃PO₄ = 0.3 M

Volume = 45.5 mL = 45.5 / 1000 = 0.0455 L

Mole = Molarity × Volume

Mole of Na₃PO₄ = 0.3 × 0.0455

Molarity of Cr(NO3)₃ = 0.2 M

Volume = 42.5 mL = 42.5 / 1000 = 0.0425 L

Mole = Molarity × Volume

Mole of Cr(NO3)₃ = 0.2 × 0.0425

Next, we shall determine the limiting reactant.

Na₃PO₄(aq) + Cr(NO₃)₃(aq) —> CrPO₄(s) + 3NaNO₃(aq)

1 mole : 1 mole

0.01365 : 0.0085 mole

Thus, Cr(NO₃)₃ is the limiting reactant.

Next, we shall determine the number of mole of CrPO₄ produced from the reaction.

From the balanced equation above,

1 mole of Cr(NO₃)₃ reacted to produce 1 mole of CrPO₄.

Therefore,

0.0085 mole of Cr(NO₃)₃ will also react to produce 0.0085 mole of CrPO₄.

Finally, we shall determine the mass of 0.0085 mole of CrPO₄.

Mole of CrPO₄ = 0.0085 mole

Molar mass of CrPO₄ = 52 + 31 + (16×4)

= 52 + 31 + 64

= 147 g/mol

Mass = mole × molar mass

Mass of CrPO₄ = 0.0085 × 147

Therefore, the mass of the precipitate, CrPO₄ formed is 1.25 g