When glucose is burned in a closed container, carbon dioxide gas and water are produced according to the following equation: $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ How many liters of \(\mathrm{CO}_{2}\) at STP can be produced when \(1.50 \mathrm{~kg}\) of glucose are burned?

Understanding how to balance chemical equations is essential in chemistry. It ensures the same number of each type of atom on both sides of the reaction. The combustion of glucose involves the following reaction:

\text{C}_6 \text{H}_{12} \text{O}_6(s) + 6 \text{O}_2(g) \rightarrow 6 \text{CO}_2(g) + 6 \text{H}_2 \text{O}(l).

In this equation, you can see that for every one molecule of glucose ( C\(_6\)H\(_{12}\)O\(_6\)), there are six molecules of oxygen ( O\(_2\)), producing six molecules of carbon dioxide ( CO\(_2\)) and six molecules of water ( H\(_2\)O). Ensuring the equation is balanced is crucial for accurate stoichiometry calculations. This helps in quantifying reactants and products in a chemical reaction accurately.

Calculating the molar mass is the next step. The molar mass of a substance is the sum of the atomic masses of all atoms in a molecule. For glucose ( C\(_6\)H\(_{12}\)O\(_6\)), the calculation is as follows:

• Carbon: 6 atoms x 12.01 g/mol = 72.06 g/mol
• Hydrogen: 12 atoms x 1.01 g/mol = 12.12 g/mol
• Oxygen: 6 atoms x 16.00 g/mol = 96.00 g/mol

Adding these values together gives the molar mass of glucose, which is 180.18 g/mol. This value is necessary when converting the mass of a substance into moles, which is pivotal for stoichiometry.

Mole-to-volume conversion is an important concept, especially when dealing with gases. At STP (Standard Temperature and Pressure), one mole of any gas occupies 22.4 liters. This is vital for converting moles of a gas into volume. Using the exercise, you've found 49.92 moles of CO2.

To convert this into volume:
Volume= Moles x 22.4 L/mol
Volume of CO2= 49.92 moles x 22.4 L/mol = 1117.2 liters.
Understanding this step ensures you can transition from theoretical quantities to real-world measurable volumes.

Stoichiometric coefficients in a chemical equation tell you the ratio of molecules taking part in the reaction. In the equation for the combustion of glucose, 1 molecule of glucose reacts with 6 molecules of oxygen to produce 6 molecules of carbon dioxide and 6 molecules of water.

The coefficient ratio helps in figuring out how many moles of one substance are involved in the reaction given the moles of another substance. For example, knowing that 1 mole of glucose yields 6 moles of CO2 tells us:

• 8.32 moles of glucose x 6 = 49.92 moles of CO2 are produced.

This ratio is foundational in all stoichiometric computations in chemistry.

The Ideal Gas Law and its applications at STP (Standard Temperature and Pressure) are essential for gas-related calculations. At STP, temperature is 273.15K (0°C) and pressure is 1 atm. A common usage is the molar volume, where 1 mole of gas equals 22.4 liters under these conditions.

Using this principle:

• Given 49.92 moles of CO2 gas, the volume can be found via Volume = Moles x 22.4 L/mol.

Thus, Volume= 49.92 moles x 22.4 L/mol = 1117.2 liters of CO2. The simplicity of this law at STP is what makes it so broadly used in introductory chemistry.