Question

Problem PageQuestion Sulfuric acid is essential to dozens of important industries from steelmaking to plastics and pharmaceuticals. More sulfuric acid is made than any other industrial chemical, and world production exceeds per year. The first step in the synthesis of sulfuric acid is usually burning solid sulfur to make sulfur dioxide gas. Suppose an engineer studying this reaction introduces of solid sulfur and of oxygen gas at into an evacuated tank. The engineer believes for the reaction at this temperature. Calculate the mass of solid sulfur he expects to be consumed when the reaction reaches equilibrium. Round your answer to significant digits. Note for advanced students: the engineer may be mistaken in his belief about the value of , and the consumption of sulfur you calculate may not be what he actually observes.

Answer 1

The question is incomplete, here is the complete question:

Sulfuric acid is essential to dozens of important industries from steel making to plastics and pharmaceuticals. More sulfuric acid is made than any other industrial chemical, and world production exceeds  2.0×10¹¹ kg per year.

The first step in the synthesis of sulfuric acid is usually burning solid sulfur to make sulfur dioxide gas. Suppose an engineer studying this reaction introduces 1.8 kg of solid sulfur and 10.0 atm of oxygen gas at 650°C  into an evacuated 50.0 L tank. The engineer believes Kp = 0.099 for the reaction at this temperature.

Calculate the mass of solid sulfur he expects to be consumed when the reaction reaches equilibrium. Round your answer to 2 significant digits.

Answer: The mass of solid sulfur that will be consumed is 19. grams

Explanation:

The chemical equation for the formation of sulfur dioxide gas follows:

                    $S(s)+O_2\rightarrow SO_2(g)$

Initial:                   10.0

At eqllm:              10-x         x

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{SO_2}}{p_{O_2}}$

We are given:

$K_p=0.099$

Putting values in above expression, we get:

$0.099=\frac{x}{10-x}\\\\x=0.901atm$

Partial pressure of sulfur dioxide = x = 0.901 atm

To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the sulfur dioxide gas = 0.901 atm

V = Volume of the gas = 50.0 L

T = Temperature of the gas = $650^oC=[650+273]K=923K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of sulfur dioxide gas = ?

Putting values in above equation, we get:

$0.901atm\times 50.0L=n\times 0.0821\text{ L. atm}mol^{-1}K^{-1}\times 923K\\\\n=\frac{0.901\times 50.0}{0.0821\times 923}=0.594mol$

By stoichiometry of the reaction:

1 mole of sulfur dioxide gas is produced from 1 mole of sulfur

So, 0.594 moles of sulfur dioxide gas will be produced from = $\frac{1}{1}\times 0.594=0.594mol$ of sulfur

• To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of sulfur = 0.594 moles

Molar mass of sulfur = 32 g/mol

Putting values in above equation, we get:

$0.594mol=\frac{\text{Mass of sulfur}}{32g/mol}\\\\\text{Mass of sulfur}=(0.594mol\times 32g/mol)=19.008g$

Hence, the mass of solid sulfur that will be consumed is 19. grams