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 !

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

m=9

step-by-step explanation:

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