A student determines the chromium (II) content of a solution by first precipitating it as chromium (II) hydroxide, and then decomposing the hydroxide to chromium (II) oxide by heating. How many grams of chromium (II) oxide should the student obtain if his solution contains 60.0 mL of 0.414 M chromium (II) nitrate

1.69 g of CrO are obtained by the student.

Cr (NO₃) ₂ (aq) → Cr²⁺ (aq) + 2NO₃⁻ (aq)

Cr²⁺ (aq) + 2OH⁻ (l) → Cr (OH) ₂ (s) ↓

Cr (OH) ₂ (s) → CrO (g) + H₂O (g) ΔH

Explanation:

The student has determined Cr²⁺

Then, he precipitated the hydroxide

Cr²⁺ + 2OH⁻ → Cr (OH) ₂ (s) ↓

Cr²⁺ comes from the Cr (NO₃) ₂

Molarity = mol / volume (L) → volume (L). Molarity = mol

We convert the volume to mL → 60 mL. 1L / 1000mL = 0.060L

0.060 L. 0.414M = 0.02484 moles of nitrate

Cr (NO₃) ₂ → Cr²⁺ + 2NO₃⁻. As ratio is 1:1, 0.02484 moles of nitrate contain 0.02484 moles of chromium ion.

We have the same amount of chromium (II) hydroxide, because ratio is 1:1, agian. The base was heated and it produced this decomposition.

Cr (OH) ₂ → CrO + H₂O

Ratio is 1:1, again. Per 1 mol of hydroxide we get 1 mol of oxide.

0.0248 moles of CrO are obtained by the student.

0.0248 mol. 68 g / 1 mol = 1.69 g