What volume of 0.400 M Hg(NO3)2 is needed to precipitate all the I- ions from 250.0 mL of 0.300 M FeI3
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To determine the volume of 0.400 M Hg(NO3)2 needed to precipitate all the I- ions from 250.0 mL of 0.300 M FeI3, we need to consider the balanced chemical equation between Hg(NO3)2 and FeI3.
The balanced chemical equation is:
2 Hg(NO3)2 + 3 FeI3 → Hg2I2 + 6 Fe(NO3)3
According to the balanced equation, 2 moles of Hg(NO3)2 react with 3 moles of FeI3 to form 1 mole of Hg2I2. This means that the molar ratio between Hg(NO3)2 and FeI3 is 2:3.
First, let's calculate the number of moles of FeI3 in the initial solution:
Moles of FeI3 = Molarity × Volume (in liters)
Moles of FeI3 = 0.300 M × 0.250 L
Moles of FeI3 = 0.075 mol
Since the molar ratio between Hg(NO3)2 and FeI3 is 2:3, we can set up a proportion to find the number of moles of Hg(NO3)2 needed:
2 mol Hg(NO3)2 / 3 mol FeI3 = X mol Hg(NO3)2 / 0.075 mol FeI3
Simplifying the proportion:
2 / 3 = X / 0.075
To find X (the number of moles of Hg(NO3)2 needed), cross-multiply:
X = (2 / 3) × 0.075
X = 0.050 mol
Now, let's determine the volume of 0.400 M Hg(NO3)2 needed to obtain 0.050 mol:
Volume (in liters) = Moles / Molarity
Volume (in liters) = 0.050 mol / 0.400 M
Volume (in liters) = 0.125 L
Finally, convert the volume to milliliters (mL):
Volume (in mL) = Volume (in liters) × 1000
Volume (in mL) = 0.125 L × 1000
Volume (in mL) = 125 mL
Therefore, 125 mL of 0.400 M Hg(NO3)2 is needed to precipitate all the I- ions from 250.0 mL of 0.300 M FeI3.