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?

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;

The actual yield of LiCl ≈ 0.035382 moles of LiCl

The theoretical yield of LiCl ≈ 0.0469475281 moles of LiCl

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

b is the correct

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