Elemental S reacts with O_2 to form SO_3 according to the reaction 2S + 3O_2 rig

Question

Elemental S reacts with O_2 to form SO_3 according to the reaction 2S + 3O_2 rightarrow 2SO_3 as depicted here.(Figure 1) How many O_2 molecules are needed to react with 2.64 g of S? Express your answer numerically in units of molecules. What is the theoretical yield of SO_3 produced by the quantities described in Part A? Express your answer numerically in grams. Elemental S reacts with O_2 to form SO_3 according to the reaction 2S + 3O_2 rightarrow 2SO_3 as depicted here (Figure 1) For each of the given situations, indicate whether S or O_2 is the limiting reactant. Drag each item to the appropriate bin.

Explanation / Answer

the reaction is :-

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

A) Given mass of sulphur = 2.64 g

moles of sulphur = 2.64/ 32  = 0.0825 mole

Two moles of sulphur reacts with three moles of oxygen

so 0.0825 moles will react with = 3 x 0.0825 / 2 = 0.124 mole

molecules of oxygen needed = 6.023 x 1023 x 0.124 = 7.47 x 1022 molecules

B) 2 moles of sulphur gives 2 moles of SO3

0.0825 moes of sulphur will give 0.0825 moles of Sulphur trioxide.

Amount of sulphur trioxide = 0.0825 x 80.07 = 6.606 g

C)Limiting reagent is Sulphur :-

3.0 Moles of Sulphur and 5.0 moles of Oxygen

Limiting reagent is Oxygen :-

3.0 moles of sulphur and 3.0 moles of oxygen

3.0 moles of sulphur and 4.0 moles of oxygen

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