Fe2O3(s) + 3CO(g) ---> 2Fe(l) + 3CO2(g) Steve inserts 450. g of iron(III) oxide and 260. g of carbon monoxide into the blast furnace. After cooling the pure liquid iron, Steve determines that he has produced 288g of iron ingots. Use the theoretical yield of liquid iron and the mass or iron ingots to calculate the percent yield of the reaction.

Theoretical yield is 313.6 g and the percent yield is, 91.8%

Explanation:

To calculate the moles :

According to stoichiometry :

1 mole of require 3 moles of

Thus 2.8 moles of will require=  of

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

As 1 mole of give = 2 moles of

Thus 2.8 moles of give = of

Mass of

Theoretical yield of liquid iron = 313.6 g

Experimental yield = 288 g

Now we have to calculate the percent yield

Therefore, the percent yield is, 91.8%

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