Iron (III) oxide is formed when Iron combines with oxygen in the air. How many grams of Fe₂O₃ are formed when 16.7 grams of reacts completely with oxygen? 4Fe + O₃ ---> 2Fe₂O₃

23.9g of Fe₂O₃ are produced

Explanation:

Are formed when 16.7g of Fe reacts completely...

Based on the reaction:

4Fe + O₃ → 2Fe₂O₃

4 moles of Iron react per 1 mole of O₃ producing 2 moles of Fe₂O₃.

To solve this question we need to convert the mass of iron to moles. The ratio of reaction is 2:1 -That is, 2 moles of Fe produce 1 mole of Fe₂O₃-. Thus, we can find the moles of Fe₂O₃ produced and its mass:

Moles Fe -Molar mass: 55.845g/mol-:

16.7g Fe * (1mol / 55.845g) = 0.299 moles of Fe

Moles Fe₂O₃:

0.299 moles Fe * (2 mol Fe₂O₃ / 4 mol Fe) = 0.150 moles Fe₂O₃

Mass Fe₂O₃ -Molar mass 159.69g/mol-:

0.150 moles Fe₂O₃ * (159.69g / mol) =

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"