Thermochemistry Pre-Lab 10

Question

A. A sample of 0.72 g of Al is reacted with 17.0 mL of 2.4 M KOH. Which one is the limiting reagent?

2Al (s) + 2KOH (aq) + 6H2O (l) → 2KAl(OH)4 (aq) + 3H2 (g)

B. Calculate the theoretical yield of KAl(OH)4 formed in question 1 if 1.0 g of Al reacts completely with KOH. Show work!

Answer 1

2Al (s) + 2KOH (aq) + 6H₂O (l) → 2KAl(OH)₄ (aq) + 3H₂ (g) ... balanced equation

A. Limiting reactant. There are a couple of ways to determine the limiting reactant. One relatively easy way is to divide the moles of each reactant by the corresponding coefficient in the balanced equation, and whichever value is less represents the limiting reactant. In this problem....

For Al: 0.72 g Al x 1 mol Al / 26.98 g = 0.02668 mole Al (÷2 -> 0.0133)

For KOH: 17.0 ml x 1 L / 1000 ml x 2.4 mol / L = 0.0336 mols KOH (÷2 -> 0.0168)

Since 0.0133 is less than 0.0168, this tells us that Al is the limiting reactant.

B. To calculate the theoretical yield if 1.0 g of Al reacts completely with excess KOH, we will use the stoichiometry of the balanced equation and at the end, convert moles to grams.

moles of Al used = 1.0 g Al x 1 mol Al / 26.98 g = 0.0371 moles Al

0.0371 mols Al x 2 mols KAl(OH)₄ / 2 mols Al = 0.0371 mols KAl(OH)₄ produced

To convert to grams, we need the molar mass of KAl(OH)4 which is

K = 39.1

Al = 27

O = 16 x 4 = 64

H = 1 x 4 = 4

Molar mass = 39 + 27 + 64 + 4 = 134 g / mole

Theoretical yield = 0.0371 mols x 134 g / mol = 4.97 g = 5.0 g (2 sig. figs.)