How many moles of lithium hydroxide would be required to produce 38.5 g of Li₂CO₃ in the following chemical reaction?

2 LiOH(s) + CO₂(g) → Li₂CO₃(s) + H₂O(l)

Answers

Answer 1
Answer:

Numer of moles of Li₂CO₃:

[tex]\longrightarrow\: n_{(Li_2 CO_3)} = \dfrac{38.5}{(6 \times 2) + 12 + (16 \times 3)}[/tex]

[tex]\longrightarrow\: n_{(Li_2 CO_3)} = \dfrac{38.5}{12+ 12 +48}[/tex]

[tex]\longrightarrow\: n_{(Li_2 CO_3)} = \dfrac{38.5}{72}[/tex]

[tex]\longrightarrow\: n_{(Li_2 CO_3)} = 0.53473 \: mol[/tex]

Chemical Reaction:

2 LiOH(s) + CO₂(g) → Li₂CO₃(s) + H₂O(l)

1 mole of Li₂CO₃ is formed from 2 moles of LiOH.

Therefore, applying unitary method:

0.53473 mole of Li₂CO₃ is formed from 2 × 0.53473 = 1.06946 moles of LiOH.
Answer 2

No. of moles of lithium hydroxide would be required to produce 38.5 g of Li₂CO₃ in the following chemical reaction are 1.03.

What is mole concept?

Avogadro's number is the number of units in one mole of any substance and equals to 6.02214076 × 10²³. The units can be electrons, atoms, ions, or molecules.

No. of moles is defined as a particular no. of particles that we can calculate with the help of Avogadro’s number.

Mass of a particular product is also find out by stoichiometry of a reaction as per the no. of mole given in the reaction.

Given,

2LiOH(s) + CO₂(g) → Li₂CO₃(s) + H₂O(l)

2 moles of LiOH are required to produce 73.8g Li₂CO₃

0.027 moles of LiOH are required to produce 1g Li₂CO₃

1.03 moles of LiOH are required to produce 38.5 Li₂CO₃

Therefore, No. of moles of lithium hydroxide would be required to produce 38.5 g of Li₂CO₃ in the following chemical reaction are 1.03.

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Answer:

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Explanation:

What’s the frequency of radiation that has a wavelength of 13 um, about the size of bacterium?

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