Question

What volume of air at 1 atm and \(273 \mathrm{~K}\) containing \(21 \%\) of oxygen by volume is required to completely burn sulphur \(\left(\mathrm{S}_{8}\right)\) present in \(200 \mathrm{~g}\) of sample, which contains \(20 \%\) inert material which does not burn. Sulphur burns according to the reaction $$ \frac{1}{8} \mathrm{~S}_{8}(s)+\mathrm{O}_{2}(g) \stackrel{\longrightarrow}{\longrightarrow} \mathrm{SO}_{2}(g) $$ (a) \(23.52\) litre (b) 320 litre (c) 112 litre (d) \(533.33\) litre

Short answer

533.33 liters

Step-by-step solution

Determine the amount of pure sulphur in the sample

Calculate the mass of pure sulphur in the sample by considering the presence of inert material. Since 20% of the sample is inert material, 80% is pure sulphur. The mass of pure sulphur is therefore 80% of 200 g, which is 0.80 * 200 g = 160 g.

Convert the mass of sulphur to moles

Use the molar mass of sulphur (S) to convert the mass of pure sulphur to moles. Since the sample is S8, the molar mass of 1 mole of S8 is 32.07 g/mol * 8 = 256.56 g/mol. The number of moles of S8 is 160 g / 256.56 g/mol = 0.624 moles of S8.

Calculate the moles of oxygen required

According to the balanced chemical reaction, 1 mole of S8 reacts with 8 moles of O2. Therefore, the moles of oxygen needed for 0.624 moles of S8 are 0.624 * 8 = 4.992 moles of O2.

Calculate the volume of oxygen at STP

At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Thus, the volume of oxygen required at STP is 4.992 moles * 22.4 liters/mole = 111.8464 liters (rounded to 112 liters).

Calculate the total volume of air required

Since air contains 21% oxygen by volume, the total volume of air needed to provide 112 liters of oxygen is 112 liters / 0.21 = 533.333 liters.

Key concepts

Physical Chemistry JEE

Physical chemistry, often featured in Joint Entrance Examination (JEE) challenges, is the study of how matter behaves on a molecular and atomic level and how chemical reactions occur. It incorporates quantitative analysis methods and requires a good understanding of mathematical concepts. In the context of JEE examinations, physical chemistry covers topics including stoichiometry, thermodynamics, equilibrium, kinetics, and many others.

For example, to solve stoichiometry problems, one must apply the principles of physical chemistry to deduce quantities such as the volume of gases involved in a reaction. When a student calculates the necessary volume of air to burn a sample based on its composition and the reaction involved, they are effectively interweaving concepts of stoichiometry, molar volumes, and gas laws—all fundamental categories within physical chemistry.

Chemical Reactions Stoichiometry

Stoichiometry is the calculation surrounding chemical reactions, involving aspects like mole ratios, mass, volume, and gaseous laws. It is a pivotal skill in accurately determining the results of chemical reactions. A stoichiometric problem, such as the one given, starts with a balanced chemical equation so that mole ratios can be established.

For example, the balanced equation \( \frac{1}{8} \mathrm{S}_{8}(s) + \mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g) \) indicates that an eighth of a mole of \(\mathrm{S}_{8}\) reacts with one mole of \(\mathrm{O}_{2}\) to produce sulfur dioxide, \(\mathrm{SO}_{2}\). Mole ratios are crucial for figuring out how much reactant is required or how much product is generated. To solve a stoichiometric problem, identifying the limiting reactant is essential—if one reactant is completely consumed, it limits the amount of product that can be formed.

Molar Mass Calculation

The molar mass is the weight of one mole of a given substance. It is a vital factor in stoichiometry, as it allows for the conversion between grams of a substance and the number of moles. Calculating molar mass requires knowledge of the atomic masses of the elements that constitute a compound, which can be found on the periodic table.

The step-by-step example demonstrates the importance of molar mass calculation. When converting the mass of pure sulfur (160 g) to moles, the molar mass of \(\mathrm{S}_{8}\) (256.56 g/mol) is used. Finding the molar mass is crucial here because it determines the number of moles of \(\mathrm{S}_{8}\), which subsequently dictates the amount of oxygen needed to burn the sulfur (given the mole ratio in the balanced equation). Without accurate molar mass calculation, it would be impossible to correctly determine the stoichiometry of the reaction.