Elemental S reacts with O2 to form SO_2 according to the reaction 2S + 3O_2 righ

Question

Elemental S reacts with O2 to form SO_2 according to the reaction 2S + 3O_2 rightarrow 2SO_3 as depicted here How many O_2 molecules are needed to react with 6.10 g of S? Express your answer numerically in units of molecules. _______________ molecules What is the theoretical yield of SO_3 produced by the quantities described in Part A? Express your answer numerically in grams. ___________ R Next consider a situation in which all of the s is consumed before all of the O_2 reacts, or one 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

Part A

The balanced chemical equation is

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

As per the stoichiometric equation,

2 moles S = 3 moles O2

Molar mass of S = 32.065 g/mol.

Moles of S corresponding to 6.10 g S = (6.10 g)/(32.065 g/mol) = 0.1902 mole.

Moles of O2 reacted = (0.1902 mole S)*(3 moles O2/ 2 moles S) = 0.2853 mole.

As per Avogadro’s hypothesis,

1 mole of O2 = 6.02*1023 molecules of O2.

Therefore, 0.2853 mole O2 = (6.02*1023 molecules/1 mole)*(0.2853 mole) = 1.7175*1023 1.72*1023 molecules of O2 (ans).

Part B

As per the balanced equation,

2 moles S = 2 moles SO3.

Therefore, 0.1902 mole S = (0.1902 mole S)*(2 moles SO3/2 moles S) = 0.1902 mole SO3.

Molar mass of SO3 = (32.065 + 3*15.9994) g/mol = 80.0632 g/mol.

Mass of SO3 produced = (0.1902 mole)*(80.0632 g/mol) = 15.2280 g 15.228 g (ans).

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