Question

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.

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 $2.0\times 10^{11}$ 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 4.4 kg of solid sulfur and 6.90 atm of oxygen gas at 950°C into an evacuated 50.0 L tank. The engineer believes $K_p=0.71$ 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 sulfur that is expected to be consumed is 0.046 kg

Explanation:

We are given:

Initial partial pressure of oxygen gas = 6.90 atm

The chemical equation for the formation of sulfur dioxide follows:

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

Initial:                  6.90  

At eqllm:             6.90-x         x

The expression of $K_p$ for above equation follows:

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

We are given:

$K_p=0.71$

Putting values in above equation, we get:

$0.71=\frac{x}{(6.9-x)}\\\\x=2.9$

So, equilibrium partial pressure of sulfur dioxide = x = 2.9 atm

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

PV = nRT

where,

P = pressure of sulfur dioxide gas = 2.9 atm

V = volume of the container = 50.0 L

n = number of moles of sulfur dioxide gas = ?

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

T = temperature of the container = $950^oC=[950+273]K=1223K$

Putting values in above equation, we get:

$2.9\times 50.0=n\times 0.0821\times 1223\\\\n=\frac{2.9\times 50.0}{0.0821\times 1223}=1.44mol$

Moles of sulfur dioxide = 1.44 moles

By Stoichiometry of the reaction:

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

So, 1.44 moles of sulfur dioxide will be produced from $\frac{1}{1}\times 1.44=1.44mol$ of sulfur

To calculate the mass of a substance, we use the equation:

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

Moles of sulfur = 1.44 moles

Molar mass of sulfur = 32 g/mol

Putting values in above equation, we get:

$1.44mol=\frac{\text{Mass of sulfur}}{32g/mol}\\\\\text{Mass of sulfur}=(1.44mol\times 32g/mol)=46.08g=0.046kg$

Hence, the mass of sulfur that is expected to be consumed is 0.046 kg