Question

The average bromine content of sea water is \(0.0064 \%\). (a) How much sea water, in cubic feet, would be required to obtain one pound of bromine? (b) What volume of chlorine gas, measured at STP, would be required to liberate the bromine from one ton of sea water? One cubic foot of sea water weighs about 63 pounds.

Short answer

(a) Approximately 248.02 cubic feet of sea water is required to obtain one pound of bromine. (b) Approximately 16.33 liters of chlorine gas, measured at STP, is required to liberate the bromine from one ton of sea water.

Step-by-step solution

Calculate the weight of bromine in one pound of sea water

To calculate the weight of bromine found in one pound of sea water, we will use the given average bromine content of sea water (0.0064%). Weight of bromine in 1 pound of sea water = 1 pound × 0.0064% Weight of bromine in 1 pound of sea water = 1 × 0.000064 pounds Weight of bromine in 1 pound of sea water = 0.000064 pounds

Calculate the weight of bromine in one cubic foot of sea water

We were given that one cubic foot of sea water weighs 63 pounds. From the previous step, the weight of bromine in one pound of sea water is 0.000064 pounds. Weight of bromine in one cubic foot of sea water = 63 pounds × 0.000064 pounds of bromine per pound of sea water Weight of bromine in one cubic foot of sea water = 0.004032 pounds

Find the volume of sea water required to obtain one pound of bromine

Now, using the result from Step 2, we can find the volume of sea water required to obtain one pound of bromine. We can set up the proportion: 1 pound of bromine = x cubic feet of sea water × 0.004032 pounds of bromine per cubic foot of sea water To solve for x: x cubic feet of sea water = 1 pound of bromine / 0.004032 pounds of bromine per cubic foot x cubic feet of sea water = 1 / 0.004032 x cubic feet of sea water = 248.02 cubic feet (approximately) (a) Therefore, approximately 248.02 cubic feet of sea water is required to obtain one pound of bromine.

Express the given weight of sea water in pounds

To find the volume of chlorine gas needed, we first need to know the weight of sea water being considered. We were given a ton of sea water to be used for this calculation. 1 ton = 2000 pounds So, the weight of sea water = 1 ton = 2000 pounds

Calculate the weight of bromine in the given weight of sea water

Using the weight of bromine in 1 pound of sea water (0.000064 pounds) obtained in Step 1, we can calculate the weight of bromine in one ton of sea water. Weight of bromine in 2000 pounds of sea water = (2000 pounds) × (0.000064 pounds of bromine per pound of sea water) Weight of bromine in 2000 pounds of sea water = 0.128 pounds

Determine the molar ratio of bromine to chlorine

In order to determine the volume of chlorine gas required to liberate bromine from sea water, we need to consider the balanced chemical equation: 2KBr + Cl₂ → 2KCl + Br₂ According to the stoichiometry of the reaction, one mole of chlorine gas (Cl₂) is needed to liberate one mole of bromine gas (Br₂) from 2 moles of potassium bromide (KBr). Molar ratio of bromine to chlorine = 1:1

Calculate the moles of bromine in the weight obtained in step 5

To calculate the moles of bromine, we will use the formula: moles of bromine = weight of bromine / molar mass of bromine The molar mass of bromine is 79.9 g/mol. moles of bromine = (0.128 pounds) * (454 g/pound) / 79.9 g/mol moles of bromine = 0.729 moles (approximately)

Calculate the moles of chlorine required

Since the molar ratio of bromine to chlorine is 1:1, the moles of chlorine required would be equal to the moles of bromine: moles of chlorine = moles of bromine moles of chlorine = 0.729 moles (approximately)

Find the volume of chlorine gas at STP

At STP (Standard Temperature and Pressure), 1 mole of any gas occupies a volume of 22.4 liters. Thus, we can find the volume of chlorine gas by multiplying the moles of chlorine by the molar volume of chlorine gas: Volume of chlorine gas = moles of chlorine × molar volume of chlorine gas Volume of chlorine gas = 0.729 moles × 22.4 L/mol Volume of chlorine gas = 16.33 L (approximately) (b) Therefore, approximately 16.33 liters of chlorine gas, measured at STP, is required to liberate the bromine from one ton of sea water.

Key concepts

Molar Mass

Understanding molar mass is crucial when dealing with chemical substances. Molar mass, expressed in grams per mole (g/mol), is the mass of one mole of a substance, where one mole contains Avogadro's number (approximately 6.022×10²³ particles) of particles, be they atoms, molecules, ions, or other entities. For instance, in our exercise, the molar mass of bromine (Br₂) is critical to determining the quantity needed for a reaction. It allows for the conversion between the mass of a substance and the number of moles, as shown in the solution steps.

For students working on similar problems, there are essential tips to accurately determine molar mass. Firstly, it's important to know the molar mass of each element, which can be found on the periodic table. For a molecule, sum the molar masses of all the atoms in the molecule. For bromine gas (Br₂), you would double the molar mass of a single bromine atom. This step in stoichiometric calculations is foundational, as errors here can cascade through subsequent calculations.

Chemical Reaction

A chemical reaction is a process where reactants are transformed into products through the breaking and forming of chemical bonds. In our exercise, we consider the reaction where potassium bromide reacts with chlorine gas to produce potassium chloride and bromine gas. Understanding the reaction's stoichiometry, which indicates the proportions of reactants and products, aids in predicting the outcomes of reactions.

To grasp chemical reactions better, students should balance chemical equations, implying that the number of atoms for each element should be equal on both sides of the reaction. This is due to the Law of Conservation of Mass, which states that mass cannot be created or destroyed in a chemical reaction. The balanced equation in our exercise indicates an equal number of bromine and chlorine molecules required for the reaction. Recognizing these ratios is vital for stoichiometric calculations and ensuring that calculations reflect the actual amounts of substances needed.

Standard Temperature and Pressure (STP)

In chemistry, Standard Temperature and Pressure (STP) is a set of conditions typically used for presenting and comparing chemical data, defined as 0°C (273.15K) and 1 atmosphere of pressure. At STP, one mole of any ideal gas occupies 22.4 liters. Knowing this is invaluable for our exercise, as we calculate the volume of chlorine gas required to react with bromine in seawater. It's a universal benchmark for comparing substances, facilitating conversions in stoichiometric calculations involving gases.

For clarity when students face challenges involving gas volumes at STP, remember that any gas's molar volume is always 22.4 L/mol at these conditions, aiding in straightforward volume calculations. Just multiply the number of moles by this molar volume to get the expected volume of gas at STP, as demonstrated in the problem's solution.

Stoichiometric Calculations

Stoichiometric calculations are methods for quantitatively analyzing the relative amounts of reactants and products in a chemical reaction. This considers the balancing of equations, understanding of molar masses, and the mole concept. In our question, we employed stoichiometry to find out how much seawater is needed to extract a pound of bromine and the volume of chlorine gas needed at STP to release bromine from a ton of seawater.

To perform stoichiometric calculations effectively, start by writing a balanced chemical equation, then use molar ratios to convert between moles of different substances. Finally, apply the molar mass and Avogadro's number as needed. Visual aids like diagrams or the 'mole map' technique can help students keep track of the steps in their stoichiometric calculations. Mastery of stoichiometry is essential for anyone progressing in chemistry, as it applies to everything from cooking recipes to industrial chemical production.