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 theor

Question

schepotkina [342]

2 years ago

15

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 theor

etically.
4Li + MnO2 ----> 2Li2O + Mn

Chemistry

1 answer:

Kazeer [188] · Answer 1 · 2023-05-23T07:31:42+03:00

Taking into account the reaction stoichiometry, 12.857 grams of Li₂O are formed when 6 grams of Li reacts.

<h3>Reaction stoichiometry</h3>

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

<h3>Mass of Li₂O formed</h3>

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?

$mass of Li_{2} O=\frac{6 grams of Lix60 grams of Li_{2} O}{28 grams of Li}$