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What mass of NaOH is required to precipitate all the Fe²+ ions from 50.0 mL of 0.200 M Fe(NO₃)₂ solution?

a) 4.00 g
b) 8.00 g
c) 16.00 g
d) 32.00 g

Answer :

Final answer:

To precipitate all the Fe²⁺ ions from a 50.0 mL of 0.200 M Fe(NO₃)₂ solution, 0.800 g of NaOH is required, calculated via stoichiometry. No provided answer choices match this result, suggesting a potential error in the answer options or lack of needed information.

Explanation:

To determine the mass of NaOH needed to precipitate all the Fe²⁺ ions from a 0.200 M Fe(NO₃)₂ solution, we should write the reaction taking place: Fe²⁺ (aq) + 2 NaOH (aq) → Fe(OH)2 (s) + 2 Na⁺ (aq) + 2 NO₃⁻ (aq)

This reaction shows that one mole of Fe²⁺ reacts with two moles of NaOH to form a precipitate of Fe(OH)2. From the reaction stoichiometry, the ratio of Fe²⁺ to NaOH is 1:2. We will use the molarity formula (molarity = moles/volume in liters) to calculate moles:

Moles of Fe²⁺ = 0.200 M × 0.050 L = 0.010 moles

Moles of NaOH required = 2 moles of NaOH per mole of Fe²⁺ × 0.010 moles Fe²⁺ = 0.020 moles NaOH

Now we calculate the mass:

Mass of NaOH = Moles of NaOH × Molar mass of NaOH
= 0.020 moles × 40.00 g/mol
= 0.800 g

Therefore, to precipitate all the Fe²⁺ ions from 50.0 mL of 0.200 M Fe(NO₃)₂ solution, 0.800 g of NaOH is required. This does not match any of the given answer options, which suggests there may be a mistake in the provided answer choices, or additional information is needed.

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