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 :

Fats and oils are used throughout the world for both food applications and industrial uses. They are consumed in butter, shortening, margarine, salad oils, and cooking oils, as well as in animal feeds, fatty acids, soaps, personal care products, biodiesel, paints (made from alkyd resins), lubricants, and greases.

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A reaction mixture initially contains 0.236 mol FeS and 0.670 mol HCl. Once the reaction has reached completion, what amount (in moles) of the excess reactant is left

Answers

Answer:

= 0.198 mole H₂S in excess

Explanation:

A quick way to determine limiting reactant is to convert reactant values given to moles and divide by respective coefficient. The smaller value is the limiting reactant other reactants will be in excess. However, when working problem, use mole values given to solve, not results of division.

       FeS         +        2HCl          => FeCl₂ + H₂S

0.236mole/1      0.670mole/2

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limiting reactant is FeS; HCl will be in excess.

Problem solution

0.236mole FeS used + 2(0.236 mole HCl used)

=> 0.236 mole FeCl₂ + 0.236 mole H₂S + (0.670 - 0.472) mole HCl in excess or, 0.198 mole H₂S in excess.

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Answer:

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Answers

Answer:

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

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