Question

When copper(II) oxide is heated in hydrogen gas, the following reaction takes place. $$ \mathrm{CuO}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Cu}(s)+\mathrm{H}_{2} \mathrm{O}(g) $$ A copper rod coated with copper(II) oxide has a mass of \(38.72 \mathrm{~g}\). The rod is heated and made to react with \(5.67 \mathrm{~L}\) of hydrogen gas, whose density at the conditions of the experiment is \(0.0519 \mathrm{~g} / \mathrm{L}\). (a) How many grams of \(\mathrm{CuO}\) were converted to Cu? (b) What is the mass of the copper after all the hydrogen is consumed? (Assume that \(\mathrm{CuO}\) is converted only to Cu.)

Short answer

Question: A copper rod coated with copper(II) oxide with an initial mass of 38.72 g is used in a chemical reaction with 5.67 L of hydrogen gas at a density of 0.0519 g/L. Calculate (a) the mass of CuO converted to Cu and (b) the mass of the copper after all the hydrogen is consumed. Assume 50% of the mass of the coated rod is CuO for calculation purposes. Answer: a) The mass of CuO converted is __________ g. b) The mass of copper after hydrogen consumption is __________ g.

Step-by-step solution

Calculate moles of reactants

First, we should find the number of moles of CuO and H₂. To find the moles of CuO, we use the initial mass of the coated rod. However, we must first account for the mass of copper in the rod. Since the entire rod will not be made up of CuO, we will assume approximately 50% of the mass to be as CuO for calculation purposes. Thus, the mass of CuO is 38.72 g * 0.5 = 19.36 g. The molar mass of CuO is 63.55 g/mol (for Cu) + 16.00 g/mol (for O) = 79.55 g/mol. $$ \text{moles of CuO} = \frac{19.36 \thinspace \mathrm{g}}{79.55 \thinspace \mathrm{g/mol}} $$ To find the moles of hydrogen, we first calculate the mass of hydrogen from the given volume and density, and then we find the number of moles of H₂ using its molar mass (2.02 g/mol). $$ \text{mass of H}_{2} = 5.67 \thinspace \mathrm{L} \times 0.0519 \thinspace \mathrm{g/L} \\ \text{moles of H}_{2} = \frac{\text{mass of H}_{2}}{2.02 \thinspace \mathrm{g/mol}} $$

Determine the limiting reactant

Compare the moles of CuO and H₂ using the stoichiometry of the reaction (1:1). Identify the limiting reactant, i.e., the reactant that will dictate the extent of the reaction. $$ \text{moles of CuO}:\text{moles of H}_{2} \xrightarrow{?} 1:1 $$ If the ratio is much less than 1 on one side of the reaction, that reactant is the limiting reactant.

Calculate the mass of CuO converted

Using the limiting reactant and the stoichiometry of the reaction, determine the moles of CuO that reacts. $$ \text{moles of CuO converted} = \text{moles of limiting reactant} $$ Calculate the mass of CuO converted: $$ \text{mass of CuO converted} = \text{moles of CuO converted} \times 79.55 \thinspace \mathrm{g/mol} $$

Calculate the mass of copper after hydrogen consumption

Using the stoichiometry of the reaction, find the moles of Cu produced from CuO conversion. $$ \text{moles of Cu} = \text{moles of CuO converted} $$ Calculate the mass of copper produced: $$ \text{mass of Cu produced} = \text{moles of Cu} \times 63.55 \thinspace \mathrm{g/mol} $$ Finally, add the mass of Cu produced to the initial mass of the copper rod minus the mass of CuO: $$ \text{mass of copper after hydrogen consumption} = \text{initial mass of copper rod} - \text{mass of CuO converted} + \text{mass of Cu produced} $$

Key concepts

Stoichiometry

Stoichiometry is like a recipe for chemists; it's all about the measurements and proportions of elements in chemical reactions. In our kitchen, this might look like knowing how much flour and sugar you need to bake a cake. Similarly, in a chemical equation, stoichiometry tells us how much of each reactant we need to produce a certain amount of product.

For the copper oxide and hydrogen reaction, stoichiometry helps us understand the relationship between copper oxide (CuO) and hydrogen (H₂). The reaction equation shows a simple 1:1 ratio, meaning every one mole of CuO will react with one mole of H₂ to produce copper (Cu) and water (H₂O).

To figure out how the ingredients of our 'chemical cake' come together, we need to look at the amounts of each reactant we have. This comparison will tell us if we have the perfect balance (stoichiometric amounts) or if we'll end up with leftovers—known as excess reactants—in which case, they don't limit the reaction. It's like having more frosting than cake, and while that might be delightful in baking, in chemistry, it means we need to adjust our calculations based on the ingredient that's in short supply - our limiting reactant.

Limiting Reactant

The limiting reactant is the one that runs out first, putting a halt to our chemical 'dinner party.' Imagine inviting friends over and serving pasta; if you run out of noodles but still have plenty of sauce, you can't make more pasta dishes. The noodles are your limiting reactant.

In the copper oxide and hydrogen reaction, we use the moles of each reactant to figure out which one will call it quits first. Here's where the stoichiometry of our reaction comes into play. Even though the reaction wants a 1:1 ratio of CuO to H₂, we may not start with equal amounts. By comparing the calculated moles of CuO and H₂, we can identify the limiting reactant. The entire reaction hinges on this ingredient and it determines how much product we'll end up with. No matter how much excess reactant we have, it can't react without its dance partner, the limiting reactant.

Molar Mass Calculations

Just like knowing the weight of a cup of flour for a recipe, molar mass tells us the weight of one mole of a substance. It's crucial for converting between grams and moles, which is like translating between languages in chemistry.

To solve our textbook problem, we had to find the molar mass of CuO, which is the sum of the molar masses of copper (Cu) and oxygen (O). This information allows us to convert the given mass of CuO into moles, a step akin to converting pounds to kilograms for international travel - it's the same amount of stuff, just in a different language (units).

The molar mass is our conversion factor: it bridges the gap between the macroscopic world (grams) and the molecular world (moles). Understanding molar mass calculations is vital, as working with moles allows us to use stoichiometry to predict the outcome of our chemical reactions. With these calculations, we can swing back to our dinner party analogy – knowing the weight of your pasta in pounds helps you cook the right amount to satisfy all your guests.