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

Answer:

87.6 %

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:

$20.0gLiOH \times \frac{1molLiOH}{23.95gLiOH} \times \frac{1molLiCl}{1molLiOH} \times \frac{42.39gLiCl}{1molLiCl} = 35.4gLiCl$

Step 3: Calculate the percent yield of LiCl.

We will use the following expression.

$\%yield = \frac{real\ yield}{theoretical\ yield} \times 100\% = \frac{31.0g}{35.4g} \times 100\% = 87.6 \%$

Answer 2

Answer:

88.6%

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:

$m_{LiCl}^{theoretical}=20.0gLiOH*\frac{1molLiOH}{23.95gLiOH}*\frac{1molLiCl}{1molLiOH} *\frac{42.394 gLiCl}{1molLiCl}=35.4gLiCl$

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

$Y=\frac{m_{LiCl}^{actual}}{m_{LiCl}^{theoretical}} *100\%=\frac{31.0g}{35.4g}*100\%\\ \\Y=87.6\%$

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

Best regards.