Question

solid iron with copper (II) sulfate. Assume the iron will form a +2 charge on the product side. Write the balanced equation for this reaction

Answer 1

hey here is the answer to your question.

Step-by-step explanation:

1) Write a balanced equation for the reaction occurring. (*Iron occurs in a few different oxidation states, but I will assume we're referring to Fe(II) or Fe2+)

Fe + CuSO4 -> Cu + FeSO4

2) Calculate the number of moles of each reactant.

Fe: 1.924 g * (1 mol / 55.85 g) = 0.03455 mol Fe

CuSO4: 0.1363 L * (0.0750 mol / 1 L) = 0.01022 mol CuSO4

3) Determine the limiting reactant. Because we have fewer moles of copper(II) sulfate and are reacting both reactants in a 1:1 ratio, we will run out of copper(II) sulfate first, making it the limiting reactant.

4) Convert moles of the limiting reactant to moles and grams of product of interest.

0.01022 CuSO4 (1 mol Cu / 1 mol CuSO4 ) (63.55 g Cu / 1 mol Cu) = 0.6495 g Cu