Question

3.26g of iron powder are added to 80.0 cm' of 0.200 moldm 3 copper(II) sulfate solution. Th
following reaction occurs:
Fe(s) + CuSO, (aq) → FeSO4 (aq) + Cu(s)

From the above equation :
Determine the limiting reactant showing your working. ​

Answer 1

Answer:

copper sulphate + iron=iron sulphide+copper

it is a displacement reaction

Explanation:

Answer 2

Answer: The limiting reagent is iron (Fe).

Given: Given mass of iron =3.26g , Volume and concentration of the solution is 80.0 cm³ and 0.200 moldm⁻³ respectively.

To find: Limiting reagent in the reaction.

Explaination:

The limiting reagent can be determined by finding the number of moles of each reactant and comparing them to the number of moles of the other reactant required for the reaction to occur.

First, we calculate the number of moles of iron powder:

mass of iron powder / molar mass of iron = 3.26 g / 55.85 g/mol = 0.058 mol

Next, we calculate the number of moles of copper(II) sulfate:

volume of solution × concentration of solution = 80.0 cm³ * 0.200 moldm⁻³ = 16.00 mol

Finally, we compare the number of moles of iron and copper(II) sulfate to the stoichiometry of the reaction:

1 mole of Fe reacts with 1 mole of CuSO₄

Since there are 0.058 mol of iron and 16.00 mol of copper(II) sulfate, we can see that the limiting reactant is iron and excess is copper(II) sulfate.

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