Question

For the following reaction, 23.1 grams of sulfur dioxide are allowed to react with 9.19 grams of oxygen gas . sulfur dioxide(g) + oxygen(g) sulfur trioxide(g)

What is the maximum mass of sulfur trioxide that can be formed? grams

What is the FORMULA for the limiting reagent?

What mass of the excess reagent remains after the reaction is complete?

Please show your work I am having a very hard time trying to understand these problems.

Answer 1

molar masses : SO2= 64, O2= 32 and SO3= 80

The reactino is SO2(g)+0.5O2(g) ------->SO3(g)

1 mole of SO2 requires 0.5 moles of oxygen to produce 1 mole of SO3. Molar ratio of SO2: O2= 1:0.5 ( theoretical ratio )

moles of gas = mass of gas/molar mass of that gas

given mass of SO2= 23.1, mole of SO2= 23.1/64=0.36, moles of oxygen =mass/molar mass=9.19/32=0.29

actual molar ratio of SO2: O3= 0.36:0.29, dividing by largest value, the ratio becomes = 0.36/0.36 :0.29/0.36= 1: 0.81

so excess reagent is Oxygen and the limiting reactant is SO2.

hence moles of SO2 reacted= moles of SO3 formed= 0.36

mass of SO3 formed =moles* molar mass = 0.36*80 =28.8 gm

moles of O2 consumed= 0.5* moles of SO2, the limiting reactant consumed =0.5*0.36= 0.18

moles of O2 remaining =0.29-0.18=0.11, mass of excess O2= 0.11*32= 3.52 gm