Answer:
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
