Question

What is the percent yield of lithium chloride if a reaction starts with 20.0 g lithium hydroxide and yields 6.00 g lithium chloride? LiOH+KCl⟶LiCl+KOH

Answer 1

Answer: The percent yield of lithium chloride if a reaction starts with 20.0 g lithium hydroxide and yields 6.00 g lithium chloride is 17.2 %

Explanation:

The balanced chemical reaction is

$LiOH+KCl\rightarrow LiCl+KOH$

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $LiOH$

$\text{Number of moles}=\frac{20.0g}{24g/mol}=0.83moles$

b) moles of $LiCl$

$\text{Number of moles}=\frac{6.00g}{42g/mol}=0.14moles$

According to stoichiometry :

1 mole of $LiOH$ produces =  1 mole of $LiCl$

Thus 0.83 moles of $LiOH$ produce=$\frac{1}{1}\times 0.83=0.83moles$ of $LiCl$

Mass of $LiCl=moles\times {\text {Molar mass}}=0.83moles\times 42g/mol=34.9g$

The theoretical yield of $LiCl$ = 34.9 g

The experimental yield of ethanol = 6.00  g

Now we have to calculate the percent yield of ethanol.

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{6.00g}{34.9g}\times 100=17.2\%$

Therefore, the percent yield is, 17.2 %