Question

Ammonium sulfide reacts with copper(II) nitrate in a double replacement reaction. What mole ratio would you use to determine the moles of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) produced if the moles of CuS are known?

Short answer

To determine the moles of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) produced, use the mole ratio of 2:1 (moles of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) to moles of CuS) from the balanced chemical equation: (NH4)2S + Cu(NO3)2 → CuS + 2NH4NO3.

Step-by-step solution

Write the balanced chemical equation

The reactants are ammonium sulfide (NH4)2S and copper(II) nitrate Cu(NO3)2. We first unbalance the chemical equation: (NH4)2S + Cu(NO3)2 → CuS + 2NH4NO3 Now we will balance it: (NH4)2S + Cu(NO3)2 → CuS + 2NH4NO3 The balanced chemical equation is: (NH4)2S + Cu(NO3)2 → CuS + 2NH4NO3

Identify the mole ratio

From the balanced chemical equation: (NH4)2S + Cu(NO3)2 → CuS + 2NH4NO3, we see that 1 mole of CuS is produced along with 2 moles of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\). So, the mole ratio of CuS to \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) is 1:2. If we are given the moles of CuS, we can determine the moles of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) produced by using the mole ratio: Moles of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) = (Moles of CuS) × (Mole ratio of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) to CuS) Mole ratio to calculate the moles of \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) produced when the moles of CuS are known is 2:1.

Key concepts

Double Replacement Reaction

When studying double replacement reactions, it's essential to recognize them as processes where two compounds exchange their components to form two new compounds. Imagine a dance where each pair decides to switch partners - this is similar to what happens at the molecular level in a double replacement reaction.

For instance, ammonium sulfide ((NH4)2S) reacts with copper(II) nitrate (Cu(NO3)2) to form copper(II) sulfide (CuS) and ammonium nitrate (NH4NO3). In simple terms, the sulfur from ammonium sulfide swaps places with the nitrate from copper(II) nitrate, leading to new partners or products. Understanding this swap helps in predicting the products of such reactions and sets the stage for further explorations of stoichiometry.

Balanced Chemical Equation

The foundational concept of a balanced chemical equation cannot be overstated. This balance is the Rosetta Stone of chemistry, providing insight into the quantities of reactants and products involved in a chemical reaction.

Here, the key principle is the conservation of mass: atoms are neither created nor destroyed during a chemical reaction, leading to the requirement that an equal number of each type of atom must be present on both sides of the equation. In the reaction of ammonium sulfide with copper(II) nitrate, it is essential to ensure that the number of ammonium, sulfur, copper, and nitrate groups remain constant. Through careful arrangement, as per the solution provided, we see a perfect balance. The equation mirrors reality, showing that one mole of ammonium sulfide reacts with one mole of copper(II) nitrate to produce one mole of copper(II) sulfide and two moles of ammonium nitrate. This step is crucial for anyone venturing into chemical calculations.

Stoichiometry

Diving into the realm of stoichiometry feels like exploring the quantitative relationships that govern the world of chemistry. This branch of chemistry is akin to a mathematical supermarket, providing the tools to calculate precisely how much of one substance reacts with another, or what quantity of a product results.

Referring to the balanced equation from our exercise, stoichiometry helps us use the mole ratio—a crucial component—to figure out the quantity of products if we know the amount of reactants. If you picture the mole as a 'chemical dozen,' it allows chemists to count atoms and molecules in a practical way, just as a baker counts eggs by the dozen. The mole ratio of CuS to NH4NO3 in our reaction is 1:2, meaning that for every one mole of CuS, two moles of NH4NO3 are produced. This simple yet profound mole ratio is the key to unlocking the amounts of substances involved. By mastering stoichiometry, students become adept at navigating the world of chemical recipes with precision.