An excess of O2 reacted with 4.60 g of Fe. What is the percent yield if...

Question

Question

An excess of O₂ reacted with 4.60 g of Fe. What is the percent yield if 6.02 g of Fe₂O₃ are isolated?

science chemistry

Answer 1

Answer #1

4Fe + 3O2 ---> 2Fe2O3

Molar mass of Fe is = 55.8 g/mol

Molar mass of Fe2O3 is = (2*55.8) + ( 3*16) = 159.6

According to the balanced Equation ,

4*55.8 g of Fe produces 2*159.6 g of Fe2O3
4.60 g of Fe produces X g of Fe2O3

X = ( 2*159.6*4.60) / (4*55.8)

= 6.578 g ---> this is the theoritical mass
Actual mass obtained is = 6.02 g

So % yield = ( actual / theoritical) * 100

= ( 6.02 / 6.578) *100

= 91.510 %

Answer 2

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