Answer: Rate of production of Sulfur dioxide is 0.15kg/s to 2 significant digits
Note: The question is incomplete. The complete question is as follows:
<em>Most of the used in the United States is chemically synthesized from hydrogen sulfide gas recovered from natural gas wells. In the first step of this synthesis, called the Claus process, hydrogen sulfide gas is reacted with dioxygen gas to produce gaseous sulfur dioxide and water Suppose a chemical engineer studying a new catalyst for the Claus reaction finds that 198. liters per second of dioxygen are consumed when the reaction is run at 186. °C and 0.69 atm. Calculate the rate at which sulfur dioxide is being produced. Give your answer in kilograms per second. Round your answer to 2 significant digits.</em>
Explanation:
The balanced equation of the reaction between hydrogen sulfide gas and dioxygen gas to produce sulfur dioxide and water is as follows:
2H₂S(g) + 3O₂(g) ---> 2SO₂(g) + 2H₂O(g)
<em>From the equation above, 3 moles of dioxide yields 2 moles of sulfur dioxide.</em>
Using the ideal gas equation to determine the number of moles of dioxygen present in 198 litres of the gas at 186. °C and 0.69 atm;
PV = nRT,
where P = 0.69atm, V = 198Litres, R (molar gas constant) = 0.082atmLK⁻¹mol⁻¹, T = 186. °C = 459K
<em>n = PV/RT</em>
n = 0.69*198*/(0.082*459)
n = 3.63moles of dioxygen
Therefore, 3.63 moles of dioxygen are consumed per second
Using the mole ratio from the equation of reaction,
<em>number of moles of sulfur dioxide produced will be 3.63moles * 2/3 = 2.42moles</em>
Therefore, the number of moles of sulfur dioxide consumed is 2.42 moles per second.
Converting to kilograms per second,
1 mole of sulfur dioxide weighs 64g (molar mass of sulfur dioxide)
<em>2.42 moles weighs 2.42*64g =154.88g or 0.15488Kg which is approximately 0.15Kg</em>
Therefore, rate of production of Sulfur dioxide is 0.15kg/s to 2 significant digits
= 0.78 mol Mg₃(Si₂O₅)₂(OH)₂