Answer :
The formula for limiting reagent is "S". The maximum amount of sulfur dioxide formed is 9.6g and the limiting reagent is 0.15 moles.
Limiting reagent is the reagent that limits or inhibit the formation of product.
As we know,
To calculate the number of moles we need to divide given mass by molar mass.
No. of moles = given mass/ molar mass.
For sulfur: -
No. of moles = 11.4 / 32
= 0.35 moles.
For carbon monoxide: -
No. of moles = 23.9 / 28
= 0.85 moles.
The chemical reaction would be S + 2 CO - SO2 + 2C
Here, we can see that 1 mole of sulfur is reacting with 2 moles of carbon monoxide. Therefore, 2/1 x 0.35 moles.
Therefore, excess amount of sulfur is 0.70.
Thus, carbon monoxide is the limiting reagent here.
For the excess amount of reagent is 0.85 - 0.70 = 0.15 moles.
By stoichiometry, we can conclude that 1 mole of sulfur gives 1 mole of sulfur dioxide.
For the excess amount of sulfur dioxide: -
Molar mass of sulfur dioxide = 64g.
Therefore, 0.15 = given mass of SO2 / 64
Mass of sulfur dioxide = 9.6 g.
For excess amount of carbon monoxide: -
Molar mass of carbon monoxide = 28g.
Therefore, 0.15 = given mass of CO / 28
Mass of carbon monoxide = 4.2 g.
Therefore, the maximum amount of sulfur dioxide produced is 9.6 g and the excess limiting reagent is 0.15 moles.
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