Question

Consider reacting copper(II) sulfate with iron. Two possible reactions can occur, as represented by the following equations. copper(II) sulfate \((a q)+\) iron \((s) \longrightarrow\) copper \((s)\) # iron(II) sulfate \((a q)\) copper \((\text { II) sulfate }(a q) \text { # iron }(s) \Longrightarrow\) copper \((s)+\) iron(III) sulfate \((a q)\) You place \(87.7 \mathrm{mL}\) of a \(0.500-M\) solution of copper(II) sulfate in a beaker. You then add \(2.00 \mathrm{g}\) of iron filings to the copper(II) sulfate solution. After one of the above reactions occurs, you isolate 2.27 g of copper. Which equation above describes the reaction that occurred? Support your answer.

Short answer

The first reaction, \( CuSO_{4}(aq) + Fe (s) \rightarrow Cu (s) + FeSO_{4}(aq) \), best describes the given reaction based on the provided information, as the calculated mass of copper from this reaction (2.273 g) is closer to the given value of 2.27 g compared to the calculated mass of copper from the second reaction (1.137 g).

Step-by-step solution

Calculate moles of copper(II) sulfate and iron

Use the volume and molarity of copper(II) sulfate solution to find the number of moles. Then, convert the mass of iron to moles. For copper(II) sulfate: Moles = Volume × Molarity = 87.7 mL × 0.500 mol/L = 0.04385 mol For iron: Moles = Mass / Molar Mass = 2.00 g / (55.85 g/mol) = 0.0358 mol

Find the limiting reactant for both reactions

Use the mole ratio to find the limiting reactant in both reactions. For the first reaction: \[ CuSO_{4}(aq) + Fe (s) \rightarrow Cu (s) + FeSO_{4}(aq) \] The mole ratio is 1:1 for copper(II) sulfate and iron. In this case, 0.04385 mol of copper(II) sulfate and 0.0358 mol of iron react. For the second reaction: \[ CuSO_{4}(aq) + 2Fe (s) \rightarrow Cu (s) + 2Fe_{2}(SO_{4})_{3}(aq) \] The mole ratio is 1:2 for copper(II) sulfate and iron. In this case, 0.04385 mol of copper(II) sulfate reacts with 0.0219 mol of iron. In both reactions, iron is the limiting reactant.

Calculate the theoretical yield of copper for both reactions

Use the mole ratio to find the amount of copper produced in both reactions. For the first reaction: Moles of copper = Moles of iron = 0.0358 mol Mass of copper = Moles × Molar Mass = 0.0358 mol × (63.55 g/mol) = 2.273 g For the second reaction: Moles of copper = 0.5 × Moles of iron = 0.0179 mol Mass of copper = Moles × Molar Mass = 0.0179 mol × (63.55 g/mol) = 1.137 g

Compare the calculated mass of copper to the given value

Compare the theoretical yield of copper from both reactions to the given value (2.27 g) to determine which reaction is most likely to have occurred. For the first reaction, the calculated mass of copper is 2.273 g, while for the second reaction, it is 1.137 g. Since 2.273 g is closer to the given value of 2.27 g, it is more likely that the first reaction occurred: \[ CuSO_{4}(aq) + Fe (s) \rightarrow Cu (s) + FeSO_{4}(aq) \] Conclusion: The first reaction best describes the given reaction based on the provided information.

Key concepts

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. In educational terms, think of it as a recipe for baking a cake, where precise amounts of ingredients are required to get the desired end product. For instance, if you have more of one ingredient (like flour) and less of another (eggs), you can't bake more cakes than your egg supply allows. Similarly, in a chemical reaction, the amount of product formed is determined by the limiting reactant, the ingredient that will run out first, just as eggs might in our cake analogy.

In the exercise provided, we examine the stoichiometry of two potential reactions between copper(II) sulfate and iron. To figure out which actual reaction occurred, we calculate the moles of each reactant, then use mole ratios to understand the balance of reactants to products. This balancing act allows us to identify the limiting reactant, akin to finding out we only have enough eggs to make one cake, not two. It's a crucial step in predicting the outcome of the reaction and amounts of each substance involved.

Theoretical Yield Calculation

When it comes to theoretical yield calculation, it's all about the 'what could be' under perfect conditions. Imagine you're planning exactly how many slices of pizza each person could get at a party based on the number of pizzas you're ordering. Theoretical yield is the maximum amount of product that can be generated from a given amount of reactants, assuming everything goes perfectly and nothing is lost in the process. It's a prediction, not the reality, because real-life reactions often encounter inefficiencies.

In the exercise, we used the limiting reactant's mole amount to calculate the theoretical yield of copper for each possible reaction. By multiplying the moles of the limiting reactant (iron in both cases) by the mole ratio from the balanced chemical equations, we determined how much copper should theoretically be produced if the entire amount of iron reacted. This step is akin to forecasting the maximum number of slices each guest can eat, without considering those who might skip the crust.

Mole-to-Mass Conversion

Mole-to-mass conversion is the bridge between the world of atoms and the world we can measure and see. It's converting back and forth between the number of particles (moles) and the mass in grams that we can put on a scale. Mole-to-mass conversion is essential because while we count atoms in moles, we typically measure substances in the lab by mass. To do this conversion, we use the substance's molar mass, which is the weight of one mole of that substance.

In our textbook problem, after finding the moles of the limiting reactant, we converted these moles to grams of copper using the molar mass of copper. This allowed us to predict the amount of copper that would be produced theoretically. In layman's terms, it's like knowing how many bags of chips (moles) you have and then figuring out the total weight of all those chips to make sure you have enough for your party-goers (mass).