Question

Suppose 8.00 g of calcium chloride are mixed with 6.00 g potassium hydroxide. What is in the container after the reaction?

The answer is below but please explain how to solve this problem.

Answer: no potassium hydroxide left; 2.86 g calcium chloride left; 3.96 g calcium hydroxide and 7.98 g potassium chloride

Answer 1

Calcium chloride has a molecular weight of 110.98 g/mol; so 8.00 g of calcium chloride will be 0.072 moles

potassium hydroxide has a moleculare weight of 56.1 g/mol; so 6.00 g of potassium hydroxide will be 0.107 moles

Reaction between calcium chloride will be

CaCl₂ + 2KOH → Ca(OH)₂ + 2KCl

so 1 mole of calcium chloride reacts with 2 moles of potassium hydroxide to give 1 mole of calcium hydroxide and 2 moles of potassium chloride

By this we have potassium hydroxide as the limiting reagent so it will be completely consumed and will in turn consume 0.107/2 moles of calcium chloride = 0.0535 moles of calcium chloride leaving

0.072 - 0.535 moles of calcium chloride unreacted which is 0.0185 moles which is 2.05 g calcium chloride

potassium chloride (MW = 74.5) formed will be 0.107 moles which is 7.97 g

Calcium hydroxide (M.W = 74.1) will be 0.0535 moles which is 3.96 g.

The value of calcium chloride in above answer is not correct. If you look for mass balance you have taken 8g + 6 g = 14g of reactant so you should be left with 14 g of product.

In the calculation I have shown the total is 2.05 + 7.97 + 3.96 which is 13.98 g due to rounding off error.

whereas in the answer above it is 2.86 + 3.96 + 7.98 = 14.8 g which is wrong.