Question

If 15.0 g calcium hydroxide (74.1 g/mol) were combined with 15.0 g hydrochloric acid (36.46 g/mol), how many grams of calcium chloride (110.98 g/mol) would form? Ca(OH)₂ + 2 HCl --> CaCl₂ + 2 H₂O.

	A.	20.0 g
	B.	none
	C.	22.8 g
	D.	22.5 g

Answer 1

Option D is correct.

Explanation.

Number of moles of Ca(OH)₂ present in 15.0 g of the same is,

                        (15.0 g) / (74.1 g/mol) = 0.202 moles

Number of moles of HCl present in 15.0 g of the same is,

                        (15.0 g) / (36.46 g/mol) = 0.411 moles

The equation for the reaction is,

Ca(OH)₂ + 2 HCl --> CaCl₂ + 2 H₂O

Hence, 1 mol of Ca(OH)₂ reacts with 2 moles of HCl to form 1 mol of CaCl₂. Therefore, 0.202 moles of Ca(OH)₂ reacts with [2 * (0.202 moles)] = 0.404 moles of HCl. Thus, Ca(OH)₂ is the limiting reagent and HCl is the excess reagent for this reaction under the given condition.

As per the stoichiometric ratio, 0.202 moles of CaCl₂ will be formed. Mass of 0.202 moles of CaCl₂ (110.98 g/mol) is,

                                    (0.202 moles) * (110.98 g/mol) = 22.418 ~ 22.5 g