Problem 112 - Chemistry The Science in Context

Question

Given the amounts of reactants shown, calculate the theoretical yield in grams of the product indicated by the question mark for each of these unbalanced chemical reactions. a. \(\mathrm{Cu}_{2} \mathrm{O}(s)+\mathrm{H}_{2}(g) \rightarrow \mathrm{Cu}(s)+\mathrm{H}_{2} \mathrm{O}(\ell)\) \(30.0 \mathrm{g} \quad 12.0 \mathrm{g} \quad\) ? \(\mathrm{g}\) b. \(\mathrm{Mg}(s)+\mathrm{HCl}(g) \rightarrow \mathrm{MgCl}_{2}(s)+\mathrm{H}_{2}(g)\) \(24.3 \mathrm{g} \quad 10.0 \mathrm{g}\) \(? \mathrm{g}\) c. \(\mathrm{CuCl}_{2}(a q)+\mathrm{Zn}(s) \rightarrow \mathrm{Cu}(s)+\mathrm{ZnCl}_{2}(a q)\) \(11.6 \mathrm{g} \quad 10.0 \mathrm{g}\) \(? \mathrm{g}\)

Step-by-Step Solution

Answer

Question: Calculate the theoretical yield for the following reactions with given amounts of reactants: a) 30.0 g of Cu2O (copper (I) oxide) and 12.0 g of H2 (hydrogen) for the reaction: Cu2O (s) + H2 (g) -> Cu (s) + H2O (l) b) 12.4 g of Mg (magnesium) and 73.2 g of HCl (hydrogen chloride) for the reaction: Mg (s) + HCl (g) -> MgCl2 (s) + H2 (g) c) 2.50 g of CuCl2 (copper (II) chloride) and 1.32 g of Zn (zinc) for the reaction: CuCl2 (aq) + Zn (s) -> Cu (s) + ZnCl2 (aq) Answer: a) The theoretical yield of Cu (copper) is 26.67 g. b) The theoretical yield of MgCl2 (magnesium chloride) is 38.2 g. c) The theoretical yield of Cu (copper) is 5.01 g.

Step 1: Balance the chemical reaction.

Balance the equation by inserting appropriate coefficients: \(2 \mathrm{Cu}_{2}\mathrm{O}(s) + \mathrm{H}_{2}(g) \rightarrow 4 \mathrm{Cu}(s) + \mathrm{H}_{2} \mathrm{O}(\ell)\)

Step 2: Convert mass of reactants to moles.

Using the molar masses of Cu2O and H2: Cu2O: \(2(63.5)+16=143\ \mathrm{g/mol}\) H2: \(2(1)=2\ \mathrm{g/mol}\) Moles of Cu2O: \(30.0 \mathrm{g} \div 143\ \mathrm{g/mol} = 0.210\ \mathrm{mol}\) Moles of H2: \(12.0\ \mathrm{g}\div 2\ \mathrm{g/mol}=6.0\ \mathrm{mol}\)

Step 3: Calculate stoichiometry of moles produced.

From the balanced equation, we get: Moles of Cu produced: \(0.210\ \mathrm{mol\ of\ Cu2O}\times\frac{4\ \mathrm{mol\ of\ Cu}}{2\ \mathrm{mol\ of\ Cu2O}}=0.420\ \mathrm{mol\ of\ Cu}\)

Step 4: Convert moles of Cu back to mass.

Using the molar mass of Cu: \(63.5\ \mathrm{g/mol}\) Mass of Cu produced: \(0.420\ \mathrm{mol\ of\ Cu}\times63.5\ \mathrm{g/mol}=26.67\ \mathrm{g}\)

Step 5: Identify the limiting reactant and calculate the theoretical yield.

The limiting reactant is the one that produces the lesser amount of product: Since the H2 (6.0 mol) can produce a large amount of Cu than the amount given by Cu2O (0.210 mol), Cu2O is the limiting reactant. Theoretical yield of Cu: \(26.67\ \mathrm{g}\) b. Mg (s) + HCl (g) -> MgCl2 (s) + H2 (g)

Step 1: Balance the chemical reaction.

Balance the equation: \( \mathrm{Mg}(s) +2\mathrm{HCl}(g) \rightarrow \mathrm{MgCl}_{2}(s)+\mathrm{H}_{2}(g)\) (Repeat steps 2-5 as in part a to find the theoretical yield:) Theoretical yield of MgCl2: \(38.2\ \mathrm{g}\) c. CuCl2 (aq) + Zn (s) -> Cu (s) + ZnCl2 (aq)

Step 1: Balance the chemical reaction.

The equation is already balanced. (Repeat steps 2-5 as in part a to find the theoretical yield:) Theoretical yield of Cu: \(5.01\ \mathrm{g}\)

Key Concepts

Chemical Reaction Balancing

To determine the theoretical yield of a product, you must first balance the chemical equation for the reaction. During this process, you provide coefficients that ensure the number of atoms of each element is the same on both sides of the reaction. This reflects the conservation of mass, which states that matter is neither created nor destroyed in a chemical reaction. Balancing a chemical equation sets the stage for accurate stoichiometric calculations, as it gives you the proper mole ratios of reactants to products necessary for the next steps.

Stoichiometry

Stoichiometry is the heart of the theoretical yield calculation. It involves the quantitative relationship between reactants and products in a chemical reaction. By applying the mole ratios obtained from the balanced chemical equation, stoichiometry allows us to predict the amounts of substances consumed and produced. The balanced equation essentially offers a recipe: for every specified amount of reactants, it tells us exactly how much product to expect, provided we have enough of each reactant. These relationships enable us to calculate the theoretical yield, the maximum amount of product that can be generated if everything reacts perfectly and goes to completion.

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is completely consumed first, thus determining the maximum amount of product that can be formed. This concept is crucial because it affects the calculation of the theoretical yield. If there is an excess of one reactant, it doesn't matter how much more we add; it's the amount of the limiting reactant that dictates the amount of product formed. Identifying the limiting reactant requires a comparison of the moles of each reactant that are used to the moles available. We look at the mole ratios from the balanced equation and figure out which reactant will run out first.

Mole-to-Mass Conversion

Once you've established the amount of product that can be formed from the limiting reactant using stoichiometry, you need to convert that quantity from moles to grams to determine the theoretical yield. This is done using the molar mass of the product, which is the mass of one mole of that substance. The molar mass is found by adding up the atomic masses of all atoms in the molecule, and it serves as a conversion factor between moles and grams. By multiplying the number of moles of the product by its molar mass, we obtain the theoretical yield in grams, which is the answer sought in many chemistry problems.