Question

Select the correct alternative from the given choices. When \(180 \mathrm{~g}\) of glucose is subjected to combustion, the volume of \(\mathrm{CO}_{2}\) liberated at \(\mathrm{STP}\) is (a) \(22.4 \ell\) (b) \(67.2\) \ell (c) \(44 \ell\) (d) \(134.4 \ell\)

Short answer

Answer: The volume of CO₂ produced is 134.4 L.

Step-by-step solution

Write the balanced chemical equation for the combustion of glucose

The combustion of glucose can be represented by the following equation: C₆H₁₂O₆ (s) + 6O₂ (g) → 6CO₂ (g) + 6H₂O (l) This equation needs to be balanced, and it already is balanced (there's already an equal number of atoms of each type on both sides).

Calculate the moles of glucose and CO₂

Given the mass of glucose as 180g, we can calculate the number of moles of glucose using its molar mass (C=12, H=1, O=16). Molar mass of glucose = 6 x 12 + 12 x 1 + 6 x 16 = 180g/mol Number of moles of glucose = mass / molar mass n(glucose) = 180g / 180g/mol = 1 mol According to the balanced chemical equation, 1 mole of glucose produces 6 moles of CO₂, so: n(CO₂) = 6 x n(glucose) n(CO₂) = 6 x 1 = 6 moles

Calculate the volume of CO₂ at STP

We know that at STP (Standard Temperature and Pressure, which is 0 °C and 1 atm), 1 mole of any gas occupies 22.4 L. Hence, we can find the volume of CO₂ by multiplying the number of moles of CO₂ by the volume occupied by 1 mole of gas at STP. Volume of CO₂ = n(CO₂) x Volume of 1 mole of gas at STP Volume of CO₂ = 6 moles x 22.4 L/mol = 134.4 L The correct answer is (d) \(134.4 \ell\).

Key concepts

Molar Mass Calculation

When delving into molar mass calculations, it's essential to understand that it's the sum of the atomic masses of all the atoms in a molecule expressed in grams per mole (g/mol). Glucose, a simple sugar with the formula \(\text{C}_6\text{H}_{12}\text{O}_6\), is a common example used in calculations like these.
To calculate the molar mass:

• Identify the atomic masses: Carbon (C) = 12 g/mol, Hydrogen (H) = 1 g/mol, and Oxygen (O) = 16 g/mol.
• Multiply each atom's molar mass by the number of times the atom appears in the molecule.
• For glucose, this results in: \(6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \text{ g/mol}\).

This calculation helps us determine how many grams make up one mole of glucose, crucial for converting between grams and moles in chemical reactions.

Stoichiometry

Stoichiometry is a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. It allows us to predict how much product will form from a given amount of reactant, based on balanced chemical equations.
For the combustion of glucose, we use the equation:

• \(\text{C}_6\text{H}_{12}\text{O}_6 (s) + 6\text{O}_2 (g) \rightarrow 6\text{CO}_2 (g) + 6\text{H}_2\text{O} (l)\). This shows that one mole of glucose reacts to produce six moles of carbon dioxide.
• This means if you start with one mole of glucose, you end up with six moles of \(\text{CO}_2\).

Using stoichiometry, you can calculate reactant and product quantities in chemical reactions, which is vital for laboratory and industrial chemical processes.

Standard Temperature and Pressure (STP)

The concept of Standard Temperature and Pressure (STP) is crucial in chemistry as it provides a common reference point for comparing gas volumes. STP is defined as a temperature of 0 degrees Celsius (273.15 Kelvin) and a pressure of 1 atmosphere (atm). At STP, one mole of an ideal gas occupies 22.4 liters.
This is particularly useful when calculating the volume of gases in reactions, as it standardizes conditions:

• In the combustion of glucose example, once you calculate the moles of \(\text{CO}_2\), you can directly multiply by 22.4 L/mol to find the gas's volume at STP.
• Knowing that six moles of \(\text{CO}_2\) are produced means that the volume at STP is \(6 \times 22.4 = 134.4\) liters.

Use of STP lets chemists work out volumes for gases without the need for complex calculations involving variable conditions of temperature and pressure.