Problem 101 - Chemistry

Question

More sulfuric acid (density \(\left.=1.8302 \mathrm{~g} / \mathrm{cm}^{3}\right)\) is produced than any other chemical-approximately \(3.6 \times 10^{11} \mathrm{lb} / \mathrm{yr}\) worldwide. What is the volume of this amount in liters?

Step-by-Step Solution

Answer

Volume of sulfuric acid is approximately \(8.918 \times 10^{10}\) liters.

Step 1: Convert Pounds to Grams

First, convert the given mass of sulfuric acid from pounds to grams. We know that 1 pound is approximately 453.592 grams. Calculate the mass in grams using: \[ 3.6 \times 10^{11} \text{ lb} \times 453.592 \text{ g/lb} = 1.6329 \times 10^{14} \text{ g} \]

Step 2: Use Density to Find Volume

Density is the mass per unit volume. Rearrange the formula for density \( \rho = \frac{m}{V} \) to solve for volume \( V \). Using the density of sulfuric acid \( \rho = 1.8302 \text{ g/cm}^3 \), calculate the volume:\[ V = \frac{m}{\rho} = \frac{1.6329 \times 10^{14} \text{ g}}{1.8302 \text{ g/cm}^3} = 8.918 \times 10^{13} \text{ cm}^3 \]

Step 3: Convert Cubic Centimeters to Liters

Finally, convert the volume from cubic centimeters to liters, knowing that 1 liter is equal to 1000 cubic centimeters:\[ V = \frac{8.918 \times 10^{13} \text{ cm}^3}{1000} = 8.918 \times 10^{10} \text{ L} \]

Key Concepts

Sulfuric Acid

Sulfuric acid is a highly important industrial chemical, largely due to its wide-ranging applications. It is used in the manufacture of fertilizers, in mineral processing, and even in wastewater processing, making it a cornerstone of modern industry. Recognized by its chemical formula \( H_2SO_4 \), this strong acid is known for its high boiling point and great affinity for water. One of the calculation aspects involves understanding its density, which is the ratio of mass to volume for a substance.
For sulfuric acid, the density is given as \( 1.8302\ \text{g/cm}^3 \). This means that every cubic centimeter of sulfuric acid weighs 1.8302 grams. When dealing with chemical calculations, especially for industrial quantities, knowing this density helps in conversions between mass and volume, crucial for logistics and storage planning.
By understanding the property of density, students grasp its integral role in scientific calculations involving liquids and ensure accurate volume conversions needed in tasks like the routine production of chemicals.

Unit Conversion

Unit conversion is a key skill in science, as real-world problems often require working with different measurement units. Here, the task involves converting mass from pounds to grams and then handling volume from cubic centimeters to liters.
The first step was to convert the mass of sulfuric acid, originally given in pounds to a more commonly used scientific unit: grams. This is done through the conversion factor \( 1\ \text{lb} = 453.592\ \text{g} \). Applying this conversion factor helps us translate the global production of sulfuric acid from pounds (\( 3.6 \times 10^{11} \ \text{lb} \)) to grams (\( 1.6329 \times 10^{14} \ \text{g} \)).
Once this mass is calculated, converting to the volume required understanding that 1 liter is equal to 1000 cubic centimeters. Unit conversions such as these ensure that all quantities are in compatible units for further calculations and analysis, a fundamental aspect of scientific accuracy.

Volume Calculation

Volume calculation in this example involves using the density formula: \( \rho = \frac{m}{V} \), where \( \rho \) is the density, \( m \) is mass, and \( V \) is volume. This equation can be rearranged to solve for volume: \( V = \frac{m}{\rho} \).
Using this formula, we can calculate the volume of sulfuric acid produced worldwide, given its mass. By substituting the mass \( 1.6329 \times 10^{14} \ \text{g} \) and the density \( 1.8302 \ \text{g/cm}^3 \), the volume comes out to be \( 8.918 \times 10^{13} \ \text{cm}^3 \).
However, typically we deal with larger units for convenience, so we convert this volume from cubic centimeters to liters, understanding that \( 1\ \text{L} = 1000\ \text{cm}^3 \). Following conversion yields \( 8.918 \times 10^{10} \ \text{L} \). This method shows how to efficiently go from knowing a material's properties to solving practical, large-scale industrial problems.