Question

2.a. 3.26 g of iron powder are added to 80.0 cm3 of 0.200 mol dm-3 copper(II)

Answer 1

Answer:

The limiting reactant is CuSO₄.

Explanation:

The reaction is:

Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)      (1)

To find the limiting reactant we need to find the number of moles of the reactants.

$\eta_{Fe} = \frac{m}{A_{r}}$

Where:

m: is the mass of iron = 3.26 g  

$A_{r}$: is the standard atomic weight of iron = 55.845 g/mol

$\eta_{Fe} = \frac{3.26 g}{55.845 g/mol} = 0.058 moles$

$\eta_{CuSO_{4}} = M*V$                

Where:

M: is the concentration of the CuSO₄ = 0.200 mol/dm⁻³

V: is the volume of the solution = 80.0 cm³

First, we need to convert the units of the volume to dm³ knowing that 1 dm = 10 cm and 1 L= 1 dm³.

$V = 80.0 cm^{3}*\frac{1 dm^{3}}{(10 cm)^{3}} = 0.080 dm^{3}$

Now, the number of CuSO₄ moles is:

$\eta_{CuSO_{4}} = M*V = 0.200 mol/dm^{3}*0.080 dm^{3} = 0.016 moles$            

So, to determine the limiting reactant we need to use the molar ratio from equation (1), Fe:CuSO₄ = 1:1

$\eta_{Fe} = \frac{1 mol Fe}{1 mol CuSO_{4}}*0.016 moles \: CuSO_{4} = 0.016 moles$

Since we need 0.016 moles of Fe to react with 0.016 moles of CuSO₄ and initially we have 0.058 moles of Fe, then the limiting reactant is CuSO₄.  

Therefore, the limiting reactant is CuSO₄.  

I hope it helps you!