Question

2.79 grams of iron is completely converted into rust. What is the weight of oxygen in the rust ( at wt fe = 55.8)

Answer 1

The reaction between iron and oxygen to form rust is as follows:


                  
                          4Fe+ 3O₂ → 2Fe₂O₃


Calculate the moles of the 2.79 grams of iron ( Fe):


Moles = mass/molar mass

          = 2.79/55.8

         =   0.05 moles

⇒We will use the mole ratio of the equation to find the moles of the rust


Mole ratio between Fe and Fe₂O₃ is 4:2 = 2:1


That means that the moles of the rust(Fe₂O₃) = 0.05/2  = 0.025moles


Calculate the mass of the Fe₂O₃ :


Mass = moles × molar mass

         = 0.025  × ( 55.8 × 2 + 16×3)

         = 0.025  × 159.6

          = 3.99

Therefore the mass of the rust formed(Fe₂O₃) is 3.99 grams

⇒Find the mass of oxygen in the rust(Fe₂O₃)

The % by mass of the oxygen 48/159.6 × 100% = 30.08%

Therefore the mass of the oxygen in the rust is:

        30.08 / 100  ×  3.99 g  = 1.2g

Therefore the mass of the oxygen in the rust formed is 1.2g

Answer 2

Answer:

1.2

Explanation:

follow the picture