How many grams of co are needed with an excess of fe2o3 to produce 35.0 g fe? show work.

We must first write out the entire equation for this reaction which is as follows:

CO + Fe2O3 --> Fe + CO2

Now we must balance this equation which provides us with the following equation:

3 CO + Fe2O3 --> 2Fe + 3 CO2

We are told that we have excess Fe2O3, so that suggests that CO is the limiting reagent. We now simply convert the mass of Fe given to moles of Fe, and convert moles of Fe to moles of CO.

35.0 g Fe/ 55.845 g/mol = 0.627 moles Fe

0.627 moles Fe x (3 moles CO)/(2 moles Fe) = 0.940 moles CO

Now with the moles of CO present, we simply convert this back to mass using the molecular weight of CO.

0.940 moles CO x 28.01 g/mol = 26.3 g CO.

Therefore, 26.3 g of CO are needed to produce 35.0 g of Fe. Since we began with three significant figures in our starting mass, our answer must also have three significant figures.

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