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Alex
3 years ago
8

A reaction starts with 20.0 grams of lithium hydroxide (LiOH) and produces 31.0 grams of lithium chloride (LiCl), what is the pe

rcent yield of lithium chloride (LiCl)? *
Chemistry
1 answer:
castortr0y [4]3 years ago
8 0

Answer:

87.6%

Explanation:

Step 1:

The balanced equation for the reaction.

LiOH + HCl —> LiCl + H2O

Step 2:

Determination of the mass of LiOH that reacted and the mass of LiCl produced from the balanced equation.

This is illustrated below:

Molar mass of LiOH = 7 + 16 + 1 = 24g/mol

Mass of LiOH from the balanced equation = 1 x 24 = 24g

Molar Mass of LiCl = 7 + 35.5 = 42.5g/mol

Mass of LiCl from the balanced equation = 1 x 42.5 = 42.5g

Thus, from the balanced equation, 24g of LiOH reacted to produce 42.5g of LiCl.

Step 3:

Determination of the theoretical yield of LiCl.

This is illustrated below:

From the balanced equation above,

24g of LiOH reacted to produce 42.5g of LiCl.

Therefore, 20g of LiOH will react to produce = (20 x 42.5)/24 = 35.4g

Therefore, the theoretical yield of LiCl is 35.4g.

Step 4:

Determination of the percentage yield of LiCl. This is illustrated below:

Actual yield of LiCl = 31g

Theoretical yield of LiCl = 35.4g

Percentage yield of LiCl =.?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 31/35.4 x 100

Percentage yield of LiCl = 87.6%

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On the basis of the Ksp values below, what is the order of the solubility from least soluble to most soluble for these compounds
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Answer:

The order of solubility is AgBr <   Ag₂CO₃ < AgCl

Explanation:

The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:

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It follows then  that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:

Comparing AgBr: Ksp = 5.4 x 10⁻¹³ with AgCl: Ksp = 1.8 x 10⁻¹⁰, AgCl will be more soluble.

Comparing Ag2CO3: Ksp = 8.0 x 10⁻¹²  with AgCl Ksp = AgCl: Ksp = 1.8 x 10⁻¹⁰ we have the complication of  the ratio of ions 2:1 in Ag2CO3,  so the answer is not obvious. But since we know that

Ag2CO3 ⇄ 2 Ag⁺ + CO₃²₋

Ksp Ag2CO3  = 2s x s = 2 s² =  8.0 x 10-12

s = 4 x 10⁻12 ∴ s= 2 x 10⁻⁶

And for AgCl

AgCl  ⇄ Ag⁺ + Cl⁻

Ksp = s² = 1.8 x 10⁻¹⁰  ∴ s = √ 1.8 x 10⁻¹⁰   = 1.3 x 10⁻⁵

Therefore, AgCl is more soluble than Ag₂CO₃

The order of solubility is AgBr <   Ag₂CO₃ < AgCl

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