A piece of iron wire weighs 2.225g. It is dissolved in acid, which oxidizes the iron to iron(II), and made up to 250 mL. A 25mL sample required 31.0 mL of a 0.0185mol/L solution of potassium dichromate. Calculate the percentage of iron in the wire.

86%

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To calculate the percentage of iron in the wire, we need to determine the number of moles of iron in the 25 mL sample and then relate it to the original iron wire's weight.

First, let's determine the number of moles of potassium dichromate used in the titration:

Molarity of potassium dichromate (K2Cr2O7) solution = 0.0185 mol/L
Volume of potassium dichromate solution used = 31.0 mL = 0.0310 L

Number of moles of K2Cr2O7 = Molarity × Volume
= 0.0185 mol/L × 0.0310 L
≈ 0.0005725 mol

Next, we need to determine the moles of iron reacted with the potassium dichromate. The balanced equation for the reaction between iron (Fe) and potassium dichromate (K2Cr2O7) is:

6 Fe + K2Cr2O7 + 7 H2SO4 → 3 Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + 7 H2O

From the equation, we can see that the mole ratio between iron and potassium dichromate is 6:1. Therefore, the number of moles of iron in the wire is:

Moles of Fe = 6 × Moles of K2Cr2O7
= 6 × 0.0005725 mol
≈ 0.003435 mol

Now, we can calculate the percentage of iron in the wire:

Percentage of iron = (Moles of Fe / Total moles in 250 mL) × 100

Total moles in 250 mL = (Molarity × Volume) of the 25 mL sample in 250 mL
= (0.0185 mol/L × 0.0250 L) / 0.250 L
≈ 0.00185 mol

Percentage of iron = (0.003435 mol / 0.00185 mol) × 100
≈ 185.68%

Therefore, the percentage of iron in the wire is approximately 185.68%.

To calculate the percentage of iron in the wire, we need to determine the number of moles of iron and the total mass of the wire.

First, let's calculate the number of moles of iron. We can use the information that a 25 mL sample required 31.0 mL of the potassium dichromate solution to determine the number of moles of iron in the 25 mL sample.

The balanced chemical equation for the reaction between iron (Fe) and potassium dichromate (K2Cr2O7) is:

6Fe2+ + Cr2O7^2- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O

From the equation, we can see that 6 moles of Fe2+ ions react with 1 mole of Cr2O7^2-. Therefore, the number of moles of Fe2+ ions in the 25 mL sample can be calculated using the following formula:

moles of Fe2+ = (volume of K2Cr2O7 solution used * concentration of K2Cr2O7 solution) / 6

Substituting the given values:

moles of Fe2+ = (31.0 mL * 0.0185 mol/L) / 6 = 0.09475 mol

Since the volume of the 25 mL sample is smaller than the total volume of the wire dissolved (250 mL), we need to scale up the number of moles of Fe2+ to represent the entire wire.

moles of Fe (in the wire) = (moles of Fe2+ in the sample * total volume of the wire) / volume of the sample

moles of Fe (in the wire) = (0.09475 mol * 250 mL) / 25 mL = 0.9475 mol

Next, let's calculate the mass of iron in the wire using the number of moles and the molar mass of iron.

The molar mass of iron (Fe) is 55.845 g/mol.

mass of iron (in the wire) = moles of Fe * molar mass of Fe
mass of iron (in the wire) = 0.9475 mol * 55.845 g/mol = 52.82894 g

Finally, we can calculate the percentage of iron in the wire using the mass of iron and the total mass of the wire.

percentage of iron = (mass of iron / total mass of the wire) * 100
percentage of iron = (52.82894 g / 2.225 g) * 100 = 2376%

Therefore, the percentage of iron in the wire is 2376%.