If you have 20.0 g of elemental sulfur (S_8) and 160. g of oxygen gas (O_2), det

Question

If you have 20.0 g of elemental sulfur (S_8) and 160. g of oxygen gas (O_2), determine which is the limiting reagent when they are combined to form sulfur dioxide (SO_2). To access a periodic table of elements, click What mass of the non-limiting reagent remains after the reaction has gone to completion? To access a periodic table of elements, click What mass of SO_2 (in grams) is formed in the completed reaction? Assume the reaction went to completion. To access a periodic table of elements, click

Explanation / Answer

The Balanced reaction is S8 + 8O2 ----> 8 SO2

Molar mass (g/mol) 256 32 64

From the balanced reaction,

1 mol = 256 g of S8 reacts with 8 mol = 8x32 =256 g of O2

20 g of S8 reacts with 20 g of O2

So 160-20 = 140 g of O2 left unreacted so O2 is the excess reactant.

Since all the mass of S8 completly reacted it is the limiting reactant

Again 1 mole = 256 g of S8 produces 8 mol = 8x64 = 512 g of SO2

20 g of S8 produces M g of SO2

M = ( 20x512)/256 = 40 g

Therefore the mass of SO2 prodced is 40 g

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