This is an oxidation-reduction (redox) reaction:
2 Mg0 - 4 e- → 2 MgII
(oxidation)
2 O0 + 4 e- → 2 O-II
(reduction)
Answer:
The yield percent is 77.68%
Explanation:
![Fe_{2}O_{3}+3CO\longrightarrow2Fe+3CO_{2}](https://tex.z-dn.net/?f=Fe_%7B2%7DO_%7B3%7D%2B3CO%5Clongrightarrow2Fe%2B3CO_%7B2%7D)
1 mole of
yields 2 moles of Fe.
So, 11.2 moles of Iron(III) oxide must yield 22.4 moles of Fe theoretically. This is according to the balanced equation.
Theoretical Yield = 22.4 moles of Fe
Experimental Yield = 17.4 moles of Fe
![Yield\:Percentage=\frac{Experimental\:Yield}{Theoretical\:Yield}\times100\\\\Yield\:Percentage=\frac{17.4}{22.4}\times100=77.68\%](https://tex.z-dn.net/?f=Yield%5C%3APercentage%3D%5Cfrac%7BExperimental%5C%3AYield%7D%7BTheoretical%5C%3AYield%7D%5Ctimes100%5C%5C%5C%5CYield%5C%3APercentage%3D%5Cfrac%7B17.4%7D%7B22.4%7D%5Ctimes100%3D77.68%5C%25)
Therefore, the yield percent is 77.68%
The answer is cumulus.
Stratus clouds often low and sometimes confused for fog. Cirrus clouds are thin and wispy.