Answer the following questions based a reaction between 80.0 g of sodium hydroxide and 90.0 g of anhydrous calcium chloride

Question

How many moles of the product compound containing calcium will be made from the given mass of sodium hydroxide?

How much of the excess reagent (in g) remains after the reaction is complete?

Assuming an experiment produced 15.0 g of the product containing calcium, what is the percent yield?

Answer 1

2NaOH(aq) + CaCl₂(aq) ==> 2NaCl(aq) + Ca(OH)₂(s) ... balanced equation

Finding the limiting reactant. One easy way is to divide the moles of each reactant by the corresponding coefficient in the balanced equation.

NaOH: 80.0 g x 1 mol / 40 g = 2 mols (÷2 -> 1)

CaCl₂: 90.0 g x 1 mol / 111 g = 0.81 mols (÷1 -> 0.81)

Limiting reactant is CaCl₂ b/c 0.81 is less than 1.

The product containing Ca is the Ca(OH)₂, with a molar mass of 74.1 g/mol

Finding the amount of Ca(OH)₂ formed:

0.81 mols Ca(OH)₂ x 1 mol Ca(OH)₂ / mol CaCl₂ x 74.1 g / mol = 60.1 g Ca(OH)₂

The excess reactant is NaOH. Amount remaining will be the initial amount less the amount use.

amount NaOH used: 0.81 mols CaCl₂ x 2 mol NaOH / 1 mol CaCl₂ = 1.62 mols

moles NaOH left over = 2 mols - 1.62 mols = 0.38 mols

mass NaOH left over = 0.38 mols NaOH x 40 g/mol = 15.2 g NaOH left

Percent yield = actual yield / theoretical yield (x100%)

% yield = 15.0 g / 60.1 g (x100%) = 30.0%