Question

A reaction starts with 20.0 g of lithium hydroxide (LiOH) and actually produces 31.0 g of lithium chloride (LiCl), what is the percent yield? (Hint: First calculate the theoretical yield of lithium chloride (LiCl))

64.5%
88.6%
81.5%
92.8%

Answer 1

87.6 %

Step-by-step explanation:

Step 1: Write the balanced equation

LiOH + KCl ⇒ LiCl + KOH

Step 2: Calculate the theoretical yield of LiCl

We will use the following relations:

• The molar mass of LiOH is 23.95 g/mol.
• The molar ratio of LiOH to LiCl is 1:1.
• The molar mass of LiCl is 42.39 g/mol.

The theoretical yield of LiCl from 20.0 g of LiOH is:

`20.0gLiOH * (1molLiOH)/(23.95gLiOH) * (1molLiCl)/(1molLiOH) * (42.39gLiCl)/(1molLiCl) = 35.4gLiCl`

Step 3: Calculate the percent yield of LiCl.

We will use the following expression.

`\%yield = (real\ yield)/(theoretical\ yield) * 100\% = (31.0g)/(35.4g) * 100\% = 87.6 \%`

Answer 2

88.6%

Step-by-step explanation:

Hello,

In this case, considering the given reaction, we notice a 1:1 molar relationship between lithium hydroxide (molar mass=23.95 g/mol) and lithium chloride (molar mass=42.394 g/mol), for that reason, we are able to compute the theoretical yield of lithium chloride by stoichiometry:

`m_(LiCl)^(theoretical)=20.0gLiOH*(1molLiOH)/(23.95gLiOH)*(1molLiCl)/(1molLiOH) *(42.394 gLiCl)/(1molLiCl)=35.4gLiCl`

Next, by knowing the actual yield of 31.0 g, we compute the percent yield as:

`Y=(m_(LiCl)^(actual))/(m_(LiCl)^(theoretical)) *100\%=(31.0g)/(35.4g)*100\%\\ \\Y=87.6\%`

Therefore, among the given, the answer should be 88.6%

Best regards.