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Taking into account the stoichiometry of the reaction, 34.12 grams of LiOH are produced from 9.89 g of Li.
In first place, the balanced reaction is:
2 Li + 2 H₂O ⇒ 2 LiOH + H₂
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= 2 moles
- H₂O= 2 moles
- LiOH= 2 moles
- H₂= 1 mole
The molar mass of each compound is:
- Li= 6.94 g/mole
- H₂O= 18 g/mole
- LiOH= 23.94 g/mole
- H₂= 2 g/mole
By reaction stoichiometry, the following amounts of mass of each compound participate in the reaction:
- Li= 2 moles× 6.94 g/mole= 13.88 g
- H₂O= 2 moles× 18 g/mole= 36 g
- LiOH= 2 moles× 23.94 g/mole= 47.88 g
- H₂= 1 mole× 2 g/mole= 2 g
Then you can apply the following rule of three: if by reaction stoichiometry 13.88 g of Li produce 47.88 g of LiOH, 9.89 g of Li produce how much mass of LiOH?
Solving:
<u><em>mass of LiOH= 34.12 grams</em></u>
Finalli, 34.12 grams of LiOH are produced from 9.89 g of Li.
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