Reaction Stoichiometry

Question

If 10.0 g of calcium hydroxide were reacted with 6.561 g of hydrochloric acid, according to the equation given :

Ca(OH) 2 (s) + HCl (aq) —

CaCl2 (aq) + H2O (1)

#1.) Which reagent acts as Limiting Reactant? Prove it with your calculations.

#2) How much calcium chloride (in grams) is formed after all the limiting reagent has been consumed?

#3.) If at the end of the reaction only 3.00 grams of water was produced, what is the percent yield of the reaction?

#4.) How much excess reagent (in grams) would remain unreacted after the completion of the reaction?

Answer 1

The equation provided IS NOT balanced. Here is the correctly balanced equation:

Ca(OH)₂(s) + 2HCl(aq) ==> CaCl₂(aq) + 2H₂O(l) .. balanced equation

(!). One way to find the limiting reactant is to simply divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever result is less indicates the limiting reactant.

For Ca(OH)₂: 10.0 g x 1 mol / 74.1 g = 0.135 mols (÷1->0.135)

For HCl: 6.561 g x 1 mol / 36.5 g = 0.1798 mols (÷2->0.089)

Since 0.089 is less than 0.135, HCl is the limiting reactant and the mols of HCl (0.1798) will determine the amount of CaCl₂ formed

(2). Mass of CaCl₂ formed = 0.1798 mols HCl x 1 mol CaCl₂ / 2 mol HCl x 111 g CaCl₂ / mol = 9.98 g CaCl₂

(3). Theoretical yield of H₂O = 0.1798 mol HCl x 2 mol H₂O / 2 mol HCl x 18 g H₂O/mol = 3.24 g H₂O

If 3.00 g is formed, the % yield is 3.00 g / 3.24 g (x100%) = 92.6% yield

(4). Excess reactant is Ca(OH)₂. To find the amount remaining, we first find out how much was used up and then subtract that from what we started with (0.135 mols)

Ca(OH)₂ used = 0.1798 mol HCl x 1 mol Ca(OH)₂ / 2 mol HCl = 0.0899 mols used

moles Ca(OH)₂ left over = 0.135 mol - 0.0899 mol = 0.0451 mols

mass Ca(OH)₂ left over = 0.0451 mols x 74.1 g / mol = 3.34 g Ca(OH)₂ left over