Question

Sulfur trioxide, ${\mathrm{SO}}_3$, is produced in enormous quantities each year for use in the synthesis of sulfuric acid. $$\begin{array}{r}\mathrm{S}(s)+{\mathrm{O}}_2(g)\to {\mathrm{SO}}_2(g)\\ 2{\mathrm{SO}}_2(g)+{\mathrm{O}}_2(g)\to 2{\mathrm{SO}}_3(g)\end{array}$$ What volume of ${\mathrm{O}}_2(g)$ at $350{.}^{\circ }\mathrm{C}$ and a pressure of $5.25\mathrm{atm}$ is needed to completely convert $5.00\mathrm{g}$ of sulfur to sulfur trioxide?

Short answer

The volume of $\mathrm{O}_{2}(g)$ needed to completely convert 5.00 g of sulfur to sulfur trioxide is 3.81 liters at 350°C and 5.25 atm.

Step-by-step solution

Write the balanced chemical equation for the reaction

The chemical equation for this reaction is given as: $$\begin{array}{r}\mathrm{S}(s)+{\mathrm{O}}_2(g)\to {\mathrm{SO}}_2(g)\\ 2{\mathrm{SO}}_2(g)+{\mathrm{O}}_2(g)\to 2{\mathrm{SO}}_3(g)\end{array}$$ By adding the two reactions, we can get the overall equation: $$\mathrm{S}(s)+{\displaystyle \frac{5}{2}}{\mathrm{O}}_2(g)\to {\mathrm{SO}}_3(g)$$

Convert the mass of sulfur to moles

Given mass of sulfur = 5.00 g Molar mass of sulfur = 32.07 g/mol To find the moles of sulfur (n_S), we use the formula: n_S = mass / molar mass $$n_S=\frac{5.00g}{32.07g/mol}$$ $$n_S=0.156mol$$

Use stoichiometry to find the moles of O₂ gas required

From the overall balanced equation, we can see that for every mole of sulfur, $\frac{5}{2}$ moles of O₂ are required. Therefore, we can find the moles of O₂ (n_O₂) required for the given mass of sulfur: $$n_{O_2}=n_S\times \frac{5}{2}$$ $$n_{O_2}=0.156\times \frac{5}{2}$$ $$n_{O_2}=0.390mol$$

Convert the moles of O₂ to volume using the Ideal Gas Law

The Ideal Gas Law formula is as follows: $$PV=nRT$$ We need to find the volume (V) of O₂ gas, and we are given the pressure (P = 5.25 atm) and the temperature (T = 350 °C). But first, we need to convert the temperature to Kelvin (K): $$T_K=T_{°C}+273.15$$ $$T_K=350+273.15$$ $$T_K=623.15K$$ Now, we can use the Ideal Gas Law to find the volume, using the gas constant R = 0.0821 (L atm)/(mol K): $$V=\frac{nRT}{P}$$ $$V=\frac{0.390\times 0.0821\times 623.15}{5.25}$$ $$V=\frac{20.00}{5.25}$$ $$V=3.81L$$ So, the volume of O₂(g) needed to completely convert 5.00 g of sulfur to sulfur trioxide is 3.81 liters at 350 °C and 5.25 atm.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a crucial concept in chemistry that helps us understand the relationship between pressure, volume, temperature, and the amount of gas in moles. The equation is expressed as:
$$PV=nRT$$
where:

• $P$ is the pressure of the gas
• $V$ is the volume
• $n$ is the number of moles
• $R$ is the ideal gas constant (0.0821 L atm/mol K)
• $T$ is the temperature in Kelvin

To find the volume of a gas, we can rearrange this formula: $$V=\frac{nRT}{P}$$
In this particular exercise, we applied the Ideal Gas Law to determine the volume of oxygen gas needed, given the pressure and temperature. The correct temperature conversion to Kelvin is vital to avoid errors, using $T_K=T_{°C}+273.15$. This transformation prepares all variables for substitution into the Ideal Gas equation, making calculations straightforward.

Stoichiometry

Stoichiometry is the area of chemistry that involves quantifying the relationships between reactants and products in chemical reactions. It's like the cookbook of chemistry, helping us to measure out precise ratios of ingredients to create desired substances. In this exercise, we focus on the transformation of sulfur to sulfur trioxide.
To convert grams of sulfur to moles (a more usable unit in reactions), we divide the mass by the molar mass: $$n_S=\frac{5.00g}{32.07g/mol}=0.156mol$$
Once we have the moles, stoichiometry allows us to determine the needed amount of oxygen—from mole ratios in the balanced equation. We find that 0.390 moles of $O_2$ are required for the reaction.

Chemical Reactions

Chemical reactions involve transforming substances through breaking and forming chemical bonds. Here, the reaction of solid sulfur with oxygen forms sulfur trioxide, a process critical in producing sulfuric acid.
The given sequential reactions show a two-step process:

• $\mathrm{S}(s)+{\mathrm{O}}_2(g)\to {\mathrm{SO}}_2(g)$
• $2{\mathrm{SO}}_2(g)+{\mathrm{O}}_2(g)\to 2{\mathrm{SO}}_3(g)$

These steps highlight how one reactant is transformed into another through the integration of oxygen. By adding these reactions, we form an overall reaction that simplifies the process to a single step: $\mathrm{S}(s)+\frac{5}{2}{\mathrm{O}}_2(g)\to {\mathrm{SO}}_3(g)$. Understanding this transformation is essential for applying stoichiometry effectively.

Balanced Chemical Equation

In chemical equations, balancing ensures the conservation of mass, meaning atoms in reactants must equal those in products. This balance is vital for accurate stoichiometry calculations.
Initially, the problem presents two half-reactions. When combined, the final balanced equation becomes:
$\mathrm{S}(s)+\frac{5}{2}{\mathrm{O}}_2(g)\to {\mathrm{SO}}_3(g)$.
This illustrates the correct ratio of reactants to form sulfur trioxide, maintaining both mass and charge conservation. By ensuring the equation is balanced, accurate relationships between reactants and products allow us to determine quantities precisely. This foundational concept aids in predicting how much of each substance is needed or produced in a chemical reaction.