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Question

1/ Use the balanced equation from number four above to determine the limiting reactant if we had a mixture of 5.0 moles of Fe and 4.0 moles of O₂.



2/ Use the balanced equation below to determine how many grams of HNO₃ can be produced if we have 20.0 grams of NO₂. (3 pts)

               3NO₂ + H₂O  -->  2HNO₃ + NO

Answer 1

1) Don't see reaction, but if we are making rust it is 4 Fe + 3 O₂ --> 2 Fe₂O₃. (5.0 mol Fe) * (2 mol Fe₂O₃ /4 mol Fe) = 2.5 mol Fe₂O₃. (4.0 mol O₂)*(2 mol Fe₂O₃ /3 mol O₂) = 2.7 mol Fe₂O₃. So Fe is limiting.

2) (20.0g NO₂) * (1 mol NO₂ /46.0g NO₂) * (2 mol HNO₃ /3 mol NO₂) * (63.0g HNO₃ /1 mol HNO₃) = 18.3g HNO₃.

Answer 2

this video will help you get the right answer

https://youtu.be/45mKzpJ5iIg