Question

\(10 \mathrm{~g}\) of carbon burns giving \(11.2\) litres of \(\mathrm{CO}_{2}\) at NTP. After combustion, the amount of unburnt carbon is (1) \(2.5 \mathrm{~g}\) (2) \(4 \mathrm{~g}\) (3) \(3 \mathrm{~g}\) (4) \(1 \mathrm{~g}\)

Short answer

4 g

Step-by-step solution

Understand the given data

Given: Mass of carbon burned = 10 g. Volume of \(\text{CO}_2\) produced = 11.2 L at NTP.

Use molar volume at NTP

At NTP, 1 mole of any gas occupies 22.4 L. Therefore, moles of \(\text{CO}_2\) are given by: \[ \text{Moles of CO}_2 = \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ moles} \]

Relate moles of \(\text{CO}_2\) to moles of carbon

The reaction for the combustion of carbon is: \[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \] One mole of carbon produces one mole of \(\text{CO}_2\). Since 0.5 moles of \(\text{CO}_2\) are produced, 0.5 moles of carbon are involved.

Calculate the mass of carbon involved

Molecular weight of carbon (C) is 12 g/mol, thus the mass of carbon involved is: \[ \text{Mass of C} = 0.5 \text{ moles} \times 12 \text{ g/mol} = 6 \text{ g} \]

Determine the unburnt carbon

Initially, 10 g of carbon was present. Out of this, 6 g was burnt. Hence, the remaining unburnt carbon is: \[ \text{Unburnt carbon} = 10 \text{ g} - 6 \text{ g} = 4 \text{ g} \]

Key concepts

molar volume

Molar volume is a crucial concept in understanding gas reactions, especially under standard conditions. At normal temperature and pressure (NTP), one mole of any gas occupies 22.4 liters. This is known as the molar volume. Knowing this, we can easily determine the number of moles of gas produced or consumed in a chemical reaction by measuring the gas’s volume.
For example, if a reaction produces 11.2 liters of carbon dioxide (CO\textsubscript{2}), we can use the molar volume to calculate the moles of CO\textsubscript{2}:
\[ \text{Moles of CO}_2 = \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ moles} \]
This means that 0.5 moles of CO\textsubscript{2} gas were produced during the reaction.

chemical reaction stoichiometry

Stoichiometry helps us understand the quantitative relationships between reactants and products in a chemical reaction. In the combustion of carbon, the balanced chemical equation is:
\[ \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \]
This equation shows that one mole of carbon (C) reacts with one mole of oxygen (O\textsubscript{2}) to produce one mole of carbon dioxide (CO\textsubscript{2}).
By knowing the moles of CO\textsubscript{2} produced (0.5 moles), we can infer the moles of carbon that reacted. Since the molar ratio of carbon to CO\textsubscript{2} is 1:1, 0.5 moles of carbon were burned during the reaction.

mass-mole conversion

Mass-mole conversions are essential in stoichiometry to relate the mass of substances to the amount in moles. The molecular weight (or molar mass) of carbon (C) is 12 grams per mole. To find out how much carbon was involved in the reaction, we convert moles of carbon to mass:
\[ \text{Mass of C} = 0.5 \text{ moles} \times 12 \text{ g/mol} = 6 \text{ g} \]
This means 6 grams of carbon reacted to produce 0.5 moles of CO\textsubscript{2}.

unburnt mass calculation

To determine how much carbon did not burn, we need to subtract the mass of carbon that reacted from the initial mass provided. Initially, we had 10 grams of carbon. From our previous calculations, we know that 6 grams of carbon were burned.
The unburnt mass is calculated as follows:
\[ \text{Unburnt carbon} = 10 \text{ g} - 6 \text{ g} = 4 \text{ g} \]
Thus, 4 grams of carbon remained unburnt after the combustion process.