Question

In the following reaction, 17.0 grams of FeS are allowed to react with 13.2 grams of O2. Which is the limiting reactant?

Answer 1

Balanced equation: 4FeS+ 30, +2F,03 + 280 Mass of FeS is 17.0g Molar mass of FeS is 87.91_8 mole Number of moles of Fes=17.0gx- 17. 0 87.919 =0.193 mole Number of moles of Fe, O, is produced = 0.193 mole FeS x- 2 mole Fe2O3 4 mole Fes =0.206 mole Fe, og Mass of O, is 13.2 g Molar mass of o, is 32.0_8 mole mole Number of moles of 0, -13.2gx 32.0g =0.413mole 2 moleFe 03 Number of moles of Fe, 0, is produced = 0.413mole 0, x= 2* 3moleo, =0.275 mole Fe, 03 FeS produces lesser number of Fe, 03. Therefore, Fes is limiting reactant.