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Sagot :
Taking into account the reaction stoichiometry, 12.857 grams of Li₂O are formed when 6 grams of Li reacts.
Reaction stoichiometry
In first place, the balanced reaction is:
4 Li + MnO₂ → 2 Li₂O + Mn
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
- Li: 4 moles
- MnO₂: 1 mole
- Li₂O: 2 moles
- Mn: 1 mole
The molar mass of the compounds is:
- Li: 7 g/mole
- MnO₂: 87 g/mole
- Li₂O: 30 g/mole
- Mn: 55 g/mole
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
- Li: 4 moles ×7 g/mole= 28 grams
- MnO₂: 1 mole ×87 g/mole= 87 grams
- Li₂O: 2 moles ×30 g/mole= 60 grams
- Mn: 1 mole ×55 g/mole= 55 grams
Mass of Li₂O formed
The following rule of three can be applied: if by reaction stoichiometry 28 grams of Li form 60 grams of Li₂O, 6 grams of Li form how much mass of Li₂O?
[tex]mass of Li_{2} O=\frac{6 grams of Lix60 grams of Li_{2} O}{28 grams of Li}[/tex]
mass of Li₂O= 12.857 grams
Finally, 12.857 grams of Li₂O are formed when 6 grams of Li reacts.
Learn more about the reaction stoichiometry:
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