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What is the percent yield of a reaction that produces 9.8 g of CO when 25.0 g of Fe304 reacts with excess carbon? Fe304 + 4C â 3Fe + 4CO O 81.096 O O 39.296 43.2% 84.8% O L incorrect 0/1 point

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Willchemistry

Hey there!:

Balanced equation for this reaction is :

Fe₃O₄ + 4 C  ----------------> 3 Fe + 4 CO

1 mole of Fe₃O₄ gives 4 moles of CO

number of moles  = mass (g ) / molar mass ( g/mol)

Molar mass of Fe₃O₄ = 231.533 g/mol

Molar mass of CO = 28.01 g/mol

Moles of Fe₃O4 = 25.0 / 231.533 => 0.108 moles

Moles of CO =  4 x 0.108 => 0.432 moles

Mass of CO = 0.432 x 28.01

Mass of CO = 12.1 g

Percent yield =  ( actual yield / theoritical yield ) x 100

Percent yield =  ( 9.8 / 12.1  ) x 100 =

Percent yield =  0.8099 x 100 =

Percent yield = 80.99 %

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