Final answer:
There is an error in the provided answer choices. The actual number of moles of electrons transferred when 511 mL of a 1.15 M NO₃– solution reacts with excess Cr(OH)₃ is not in the choices provided; the calculated value is 1.76145 moles.
Step-by-step explanation:
When 511 mL of a solution that is 1.15 M in NO₃– reacts with excess Cr(OH)₃ solid, we need to determine the number of moles of electrons transferred during the reaction. To answer this, we can write the half-reaction for chromium going from Cr(OH)₃ to CrO₄²⁻:
H₂O(l) + Cr(OH)₃ (s) → CrO4²⁻(aq) + 5H⁺ (aq) + 3e⁻
From this half-reaction, we see that for every mole of Cr(OH)₃ that reacts, 3 moles of electrons are transferred. To find the total number of moles of electrons transferred, we first calculate the moles of NO₃⁻ in the solution using the concentration (Molarity) and the volume of the solution:
Moles of NO₃⁻ = 1.15 moles/L × 0.511 L = 0.58715 moles
The reaction stoichiometry between NO₃⁻ and Cr(OH)₃ is 3:1, which means the moles of Cr(OH)₃ reacting is equal to the moles of NO₃⁻. Moles of electrons would be 3 times the moles of Cr(OH)₃ reacting. Therefore, we have:
Moles of electrons transferred = 3 x 0.58715 moles = 1.76145 moles
Since none of the answer choices closely match the calculated value, it's likely that there is an error in the provided choices, or an error in the original question. However, the calculation process for the number of moles of electrons transferred in a reaction such as this is demonstrated correctly.