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How many grams of carbon monoxide are needed to react with an excess of iron (III) oxide to produce 198.5 grams of iron? Fe2O3(s) + 3CO(g) -----> 3CO2(g) + 2Fe(s) Show all work step by step please!

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Fe2O3(s) + 3CO(g) -----> 3CO2(g) + 2Fe(s)


mass of iron = 198.5 g of iron

atomic mass of iron = 55.845g/mol


moles of iron = mass/Atomic mass

= 198.5g/55.845

= 3.5545 moles Fe


moles of carbon monoxide

= 3.5545 moles Fe x 3 moles CO

2 moles Fe

= 5.3317 moles CO


molar mass of CO = 12.01 + 16 = 28.01g/mol


mass of carbon monoxide reacted

= moles x molar mass

= 5.3317 x 28.01

= 149.34g


grams of carbon monoxide needed to react with an excess of iron(II)oxide to produce 198.5g of iron is 149.34g CO

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