I am looking for some help with Chemistry specificaly limiting reagent and stoic

Question

I am looking for some help with Chemistry specificaly limiting reagent and stoiciometry

I need a step by step response please.

3Fe2O3+CO--->2Fe3O4+CO2

If 125g of Fe2O3 and 125 g of CO react in this first step, determine the limiting reagent and the mass in grams of each product formed. Also determine the mass in grams of excess reagent that reacts and the mass in grams of excess reagent that remains after the reaction is complete.

Here are some calculations already done for you to help you get it done faster

Molar masses.

Fe2O3=159.7g CO=28.01g Fe3O4=231.5g CO2=44.01

Explanation / Answer

                           Given masses               Molar Mass

Fe2O3    125 g                        159.7 g

CO                             125 g                        28.01 g

Fe3O4                           ?                            231.5 g

CO2                              ?                            44.01 g

Number of moles = Mass / Molar mass

Moles of 3 x Fe2O4 = 3 x 125/159.07 = 3 x 0.785 = 2.355 moles

Moles of CO = 125/28.01 = 4.462 moles

*Since CO is in abundance, Fe2O3 is the limiting reagent.

So, 2 x 0.785 = 1.57 moles of Fe3O4

and 0.785 moles of CO2 are produced

Masses of products:

Mass = number of moles x molar mass

Mass of Fe3O4 = 1.57 x 231.5 = 363.4 g

Mass of CO2 = 0.785 x 44.01 = 34.5 g

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