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If you have 6.0-g of lithium and you add it to excess manganese (IV) oxide, how many grams of the product Li2O do you form theoretically.
4Li + MnO2 ----> 2Li2O + Mn


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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