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Anit [1.1K]
3 years ago
9

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.
Chemistry
1 answer:
baherus [9]3 years ago
6 0

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

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} Fe_2O_3=\frac{450}{160}=2.8moles

\text{Moles of} CO=\frac{260}{28}=9.3moles

Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(l)+3CO_2(g)

According to stoichiometry :

1 mole of Fe_2O_3 require 3 moles of CO

Thus 2.8 moles of Fe_2O_3 will require=\frac{3}{1}\times 2.8=8.4moles  of CO

Thus Fe_2O_3 is the limiting reagent as it limits the formation of product and CO is the excess reagent.

As 1 mole of Fe_2O_3 give = 2 moles of Fe

Thus 2.8 moles of Fe_2O_3 give =\frac{2}{1}\times 2.8=5.6moles of Fe

Mass of Fe=moles\times {\text {Molar mass}}=2.6moles\times 56g/mol=313.6g

Theoretical yield of liquid iron = 313.6 g

Experimental yield = 288 g

Now we have to calculate the percent yield

\%\text{ yield}=\frac{\text{Actual yield }}{\text{Theoretical yield}}\times 100=\frac{288g}{313.6g}\times 100=91.8\%

Therefore, the percent yield is, 91.8%

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