Question

A solution contains cr3+ ions. the addition of 0.063 l of 1.50 m naf solution was needed to completely precipitate the chromium ions as crf3(s). what was the mass of cr3+ ions in the original solution?

Answer 1

Cr{3+} + 3 NaF → CrF3 + 3 Na{+} 

First calculate the total mols of NaF. 

(0.063 L) x (1.50 mol/L NaF) = 0.0945 mol NaF total 

 

Using stoichiometric ratio:

0.0945 mol NaF * (1 mol Cr3+ / 3 mol NaF) * (51.9961 g Cr3+/mol) = 1.6379 g Cr3+

Answer 2

The mass of chromium ion, Cr³⁺, in the original solution is 1.638 g

• We'll begin by calculating the number of mole of NaF in the solution. This can be obtained as follow:

Volume = 0.063 L

Molarity of NaF = 1.50 M

Mole of NaF =?

Mole = Molarity x Volume

Mole of NaF = 1.5 ×0.063

Mole of NaF = 0.0945 mole

• Next, we shall determine the mole of Cr³⁺ needed to react with 0.0945 mole of NaF. This is given below:

Cr³⁺ + 3NaF —> CrF₃ + 3Na⁺

From the balanced equation above,

3 moles of NaF reacted with 1 mole of Cr³⁺.

Therefore,

0.0945 mole of NaF will react with = $\frac{0.0945}{3}$ = 0.0315 mole of Cr³⁺.

• Finally, we shall determine the mass of 0.0315 mole of Cr³⁺.

Mole of Cr³⁺ = 0.0315 mole

Molar mass of Cr³⁺ = 52 g/mol

Mass of Cr³⁺ =?

Mass = Mole × molar mass

Mass of Cr³⁺ = 0.0315 × 52

Mass of Cr³⁺ = 1.638 g

Thus, the mass of chromium ion, Cr³⁺, in the original solution is 1.638 g

Learn more: https://brainly.com/question/22387155