Question

Lithium reacts spontaneously with bromine to produce lithium bromide. Write the balanced chemical equation for the reaction. If 25.0 g of lithium and 25.0 g of bromine are present at the beginning of the reaction, determine a. the limiting reactant. b. the mass of lithium bromide produced. c. the excess reactant and the excess mass.

Short answer

a. The limiting reactant is bromine (Br2). b. The mass of lithium bromide (LiBr) produced is approximately 27.1 grams. c. The excess reactant is lithium (Li) with an excess mass of approximately 22.8 grams.

Step-by-step solution

Write the balanced chemical equation

For the reaction between lithium (Li) and bromine (Br2) to produce lithium bromide (LiBr), we write the chemical equation: \( \) Li + Br2 -> LiBr \( \) Now we must balance the equation. As it stands, there are two bromine atoms on the left side and only one on the right side. To balance the equation, we should add a coefficient of 2 in front of LiBr on the right side of the equation: \( \) 2Li + Br2 -> 2LiBr \( \) Now the balanced chemical equation is: 2Li + Br2 -> 2LiBr.

Find the limiting reactant

We're given 25.0 g of lithium and 25.0 g of bromine. First, we need to determine the number of moles for each reactant. The molar mass of lithium (Li) is approximately 6.94 g/mol, and the molar mass of bromine (Br2) is approximately 159.8 g/mol. Moles of Li = (25.0 g Li) / (6.94 g/mol) ≈ 3.60 mol Li Moles of Br2 = (25.0 g Br2) / (159.8 g/mol) ≈ 0.156 mol Br2 Next, we divide the moles of each reactant by their respective stoichiometric coefficient from the balanced chemical equation to find the limiting reactant. Li = 3.60 mol / 2 ≈ 1.80 Br2 = 0.156 mol / 1 ≈ 0.156 Since the value for Br2 is lower, bromine is the limiting reactant.

Calculate the mass of lithium bromide produced

Using the limiting reactant, Br2, we can now calculate the mass of lithium bromide (LiBr) produced in the reaction. Moles of LiBr produced = (moles of Br2) x (2 moles LiBr/1 mole Br2) = 0.156 mol Br2 * (2/1) = 0.312 mol LiBr We can convert moles of LiBr to grams using its molar mass (approximately 86.85 g/mol). Mass of LiBr produced = (0.312 mol LiBr) x (86.85 g/mol) ≈ 27.1 g Thus, the mass of lithium bromide produced is approximately 27.1 grams.

Determine the excess reactant and its excess mass

Since we already established that bromine (Br2) is the limiting reactant, lithium (Li) would be the excess reactant. Now, we must determine the excess mass of lithium remaining after the reaction is complete. Moles of Li consumed = (moles of Br2) x (2 moles Li/1 mole Br2) = 0.156 mol Br2 * (2/1) = 0.312 mol Li Initially, we had 3.60 mol of Li, and 0.312 mol was consumed in the reaction: Remaining moles of Li = 3.60 mol - 0.312 mol = 3.29 mol Excess mass of Li = (3.29 mol Li) x (6.94 g/mol) ≈ 22.8 g Thus, the excess reactant is lithium (Li) and the excess mass is approximately 22.8 grams.

Key concepts

Limiting Reactant

In chemical reactions, the limiting reactant is the substance that runs out first and thus limits the amount of product that can be formed. It's like the smallest pack of sugar in a pancake mix; once it's finished, you can't make any more pancakes!

To determine which reactant is limiting, we compare the amount of each substance we have to its stoichiometric ratio in the balanced chemical equation. By calculating the moles of each reactant and dividing them by their respective coefficients from the balanced equation, we find the reactant that produces the least amount of product. This is the limiting reactant.

In our example, bromine (Br2) was identified as the limiting reactant because, after adjusting for stoichiometry, it produces less product than lithium (Li). Thus, the reaction stops once all the bromine is consumed.

Balanced Chemical Equation

A balanced chemical equation is essential in chemistry because it provides the correct proportions of reactants and products. It respects the law of conservation of mass, ensuring that atoms are neither created nor destroyed in a reaction.

Balancing involves adding coefficients to the chemical formulas to equalize the number of atoms for each element on both sides of the equation. For example, in the reaction of lithium with bromine, we had:

- Unbalanced: Li + Br2 → LiBr

- Balanced: 2Li + Br2 → 2LiBr

By balancing, we ensure the ratio matches what actually occurs, allowing us to accurately calculate the quantities of reactants and products involved.

Moles Calculation

Moles are the bridge between the atomic world and the world we see. They allow us to count atoms by weighing them, thanks to Avogadro's number. Calculating moles helps us determine how much of a substance is involved in a reaction.

To compute moles from a given mass, we use the formula: \( \text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \). For example, with 25.0 g of lithium (Li), and knowing its molar mass is 6.94 g/mol, we found:

\( \frac{25.0 \text{ g Li}}{6.94 \text{ g/mol}} \approx 3.60 \text{ mol Li} \).

This calculation ensures that we take into account the correct amount of reactants when considering the reaction's stoichiometry.

Excess Reactant

The excess reactant is the opposite of the limiting reactant. It's the reactant that remains after a chemical reaction has stopped. Think of it as the extra flour left over after making a batch of cookies.

In scenarios where calculations are based on the limiting reactant, the excess reactant is what is leftover. Knowing the excess reactant allows us to determine how much of it remains unused.

In the exercise, lithium (Li) was the excess reactant. After the reaction consumed 0.312 moles out of the initial 3.60 moles of Li, there were still 3.29 moles left.

Converting this back to mass, we found approximately 22.8 grams of lithium was in excess, showcasing that it's important to manage resources efficiently.