Elemental S reacts with O2 to form SO3 according to the reaction 2S+3O22SO3. Par

Question

Elemental S reacts with O2 to form SO3 according to the reaction 2S+3O22SO3.

Part A How many O2 molecules are needed to react with 5.69 g of S? Express your answer numerically in units of molecules.

Part B What is the theoretical yield of SO3 produced by the quantities described in Part A? Express your answer numerically in grams.

Limiting reactant

Next, consider a situation in which all of the S is consumed before all of the O2 reacts, or one in which you have excess S because all of the O2 has been used up.

Part C For each of the given situations, indicate whether S or O2 is the limiting reactant.

Drag each item to the appropriate bin.

3.0 mol sulfur 3.0 mol oxygen 3.0 mol sulfur 4.0 mol oxygen 3.0 mol sulfur 5.0 mol oxygen

Bin 1: Limiting reactant is sulfur

Bin 2: Limiting reactant is oxygen

Explanation / Answer

A)

2S + 3 O2 -------------------> 2 SO3

5.69 g 'S' = 5.69 g / 32 g/ mol = 0.1778 mole

2 mole S = 3 mole O2

0.1778 mole S = 0.2667 mole O2

                             x 6.023 x 10 23 molecule /mole

                      -----------------------------------------------------------

                        1.606 x 10 23 molecule of O2

                     --------------------------------------------------------------------

B) 2 mole S = 2 Mole SO3

0.1778 mole S = 0.1778 mole SO3

                              x 80 g / mole

                  -------------------------------------------------

                        14.224 g of SO3

                 ----------------------------------------------------

C)

      3 mol S require 4.5 mole of O2

so case1 and case 2 are deficient of O2 (so here limiting reagent is O2)----------added to " BIN 2 "

In case 3 is deficient of S (so here limiting reagent is S)----------added to " BIN 1 "

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