5.0g of impure iron reacted completely with 75cm3 of 2.0 mlodm-3 hydrochloric acid to give a neutral solution

Calculate the percentage purity of the sample of iron

1 answer

To calculate the percentage purity of the sample of iron, we need to determine how much pure iron is contained in the impure sample when it reacts with hydrochloric acid (HCl). The stoichiometry of the reaction and the reaction conditions will guide our calculations.

Step 1: Determine the moles of hydrochloric acid

First, we will calculate the number of moles of hydrochloric acid that reacted.

The concentration of hydrochloric acid is given as 2.0 mol/dm³, and the volume is given as 75 cm³. We need to convert the volume from cm³ to dm³:

\[ \text{Volume} = 75 , \text{cm}^3 = 75 \times 10^{-3} , \text{dm}^3 = 0.075 , \text{dm}^3 \]

Now we can calculate the number of moles of HCl:

\[ \text{Moles of HCl} = \text{Concentration} \times \text{Volume} = 2.0 , \text{mol/dm}^3 \times 0.075 , \text{dm}^3 = 0.15 , \text{moles} \]

Step 2: Write and analyze the chemical reaction

The reaction of iron with hydrochloric acid can be represented as:

\[ \text{Fe(s)} + 2 \text{HCl(aq)} \rightarrow \text{FeCl}_2(aq) + \text{H}_2(g) \]

From the balanced equation, we see that 1 mole of iron reacts with 2 moles of hydrochloric acid.

Step 3: Determine moles of iron that reacted

Given that the number of moles of hydrochloric acid is 0.15 moles, we can determine the number of moles of iron that reacted:

\[ \text{Moles of Fe} = \frac{\text{Moles of HCl}}{2} = \frac{0.15}{2} = 0.075 , \text{moles} \]

Step 4: Convert moles of iron to mass

The molar mass of iron (Fe) is approximately 55.85 g/mol. Now, we can calculate the mass of pure iron that reacted:

\[ \text{Mass of Fe} = \text{Moles of Fe} \times \text{Molar Mass of Fe} = 0.075 , \text{moles} \times 55.85 , \text{g/mol} \approx 4.19 , \text{g} \]

Step 5: Calculate percentage purity

Finally, we can calculate the percentage purity of the sample of iron:

\[ \text{Percentage Purity} = \left( \frac{\text{mass of pure Fe}}{\text{mass of impure sample}} \right) \times 100 = \left( \frac{4.19 , \text{g}}{5.0 , \text{g}} \right) \times 100 \approx 83.8% \]

Conclusion

The percentage purity of the sample of iron is approximately 83.8%.