Question

For the following reaction, 23.2 grams of sulfur dioxide are allowed to react with 4.96 grams of oxygen gas. sulfur dioxide (g) + oxygen (g)- sulfur trioxide (g) What is the maximum amount of sulfur trioxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams Submit Answer Retry Entire Group 9 more group attempts remaining

Answer 1

The balanced equation for this reaction is :-

2SO2 + O2 ------> 2SO3

Лаоко # Sulfur dionice og tonggeng, sulfur 250₂ +4 280₂. 23.29 4.969 of So₂ = 23.2 = 0.362 mole 64.066 of 2 = 4.96 = 0.155 mole #6 32 2 mole SO₂ react with Imole o 0.362 mole so₂ = 1 x 0.362 moleo, = 0.181 mole Oz .. limiting Reagent is 0 155 mdeo will react with 0.31mde so₂ I mole O₂ gives a mole SO3 0.155 mole o2 = 0.31 mole SO₃ Amount of S03 formed = 0.31x80.066 og = 24.82g Excess Reagent = 502 Moles of so ₂ reacted = 0.31 mole I unreacted = 0.362 -0.31 = 0.052 Amount of SO2 unreacted = 0.052 x 64.066 =3.331g