Question

LiOH + KCl ----> LiCl + KOH 5.00 grams of LiOH react with 3.50 grams of KCl and 1.50 grams of LiCl are produced. What is the precent yield of LiCl?

Answer 1

Answer:

The percentage yield of LiCl is approximately 75.365%

Explanation:

The given chemical reaction is presented as follows;

LiOH + KCl → LiCl + KOH

1 ole of LiOH reacts with 1 mole of KCl to produce 1 mole of LiCl and 1 mole of KOH

The mass of LiOH in the reaction = 5.00 grams

The mass of KCl in the reaction = 3.50 grams

The mass of LiCl produced in the reaction = 1.50 grams

The number of moles, n = Mass/(Molar mass)

The molar mass of LiOH = 23.95 g/mol

The number of moles of LiOH in the reaction, n = 5.00 g/(23.95 g/mol) ≈ 0.209 moles

The molar mass of KCl = 74.5513 g/mol

The number of moles of KCl in the reaction, n = 3.50 g/(74.5513 g/mol) ≈ 0.0469475281 moles

The molar mass of LiCl = 42.394 g/mol

The number of moles of LiCl produced, n = 1.50 g/(42.394 g/mol) ≈ 0.035382 moles

Therefore, the limiting reactant = KCl with 0.0469475281 moles

We get;

0.0469475281 moles of LiOH should react with 0.0469475281 moles of KCl to procude 0.0469475281 moles of LiCl and 0.0469475281 moles of KOH

The percentage yield is given as follows;

$Percentagee \ yield = \dfrac{Actual \ yield}{Theoretical \ yield } \times 100$

The actual yield of LiCl ≈ 0.035382 moles of LiCl

The theoretical yield of LiCl ≈ 0.0469475281 moles of LiCl

$\therefore The \ percentage \ yield \ of \ LiCl = \dfrac{0.035382 }{0.0469475281 } \times 100 \approx 75.365 \%$

The percentage yield of lithium chloride, LiCl ≈ 75.365%.