Mining companies extract iron from iron ore according to the following balanced equation: Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g) In a reaction mixture containing 175 g Fe2O3 and 73.0 g CO, CO is the limiting reactant. Calculate the mass of the reactant in excess (which is Fe2O3) that remains after the reaction has gone to completion. Express the mass with the appropriate units.

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thomasdecabooter thomasdecabooter · Answer 1 · 2019-12-11T21:10:28+01:00

Answer:

There will remain 36.7 grams of Fe2O3

Explanation:

Step 1: Data given

Mass of Fe2O3 = 175 grams

Mass of CO = 73.0 grams

CO is the limiting reactant

Molar mass of CO = 28 g/mol

Molar mass of Fe2O3 = 159.69 g/mol

Step 2: The balanced equation

Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)

Step 3: Calculate moles of CO

Moles CO = mass CO / molar mass CO

Moles CO = 73.0 grams / 28.0 g/mol

Moles CO = 2.61 moles

Step 4: Calculate moles of Fe2O3

Moles Fe2O3 = 175.0 grams / 159.69 g/mol

moles Fe2O3 = 1.10 moles

Step 5: Calculate moles reacted

Co is the limiting reactant. It will completely be consumed (2.61 moles)

Fe2O3 is in excess. There will be 2.61/3 = 0.87 moles of Fe2O3 consumed

There will remain 1.10 - 0.87 = 0.23 moles of Fe2O3

Step 6: Calculate the mass of Fe2O3 that remains

Mass Fe2O3 = moles * molar mass

Mass Fe2O3 = 0.23 moles * 159.69 g/mol

Mass of Fe2O3 = 36.7 grams

There will remain 36.7 grams of Fe2O3