calculate how many grams of Cr (OH) 3 are formed when 19 kg of KOH reacts with CrCl3

Question

calculate how many grams of Cr (OH) 3 are formed when 19 kg of KOH reacts with CrCl3

Answer 1

Hello, Marcuss,

We need to start with a balanced equation. I arrive at:

CrCl₃ + 3KOH = Cr(OH)₃ + 3 KCL

We are given 19 grams of KOH. We'll assume there is enogh CrCl₃ to react with all of the potassium hydroxide.

The balanced equation states that we should expect 1 mole of Cr(OH)₃ for every 3 moles of KOH:

(1 mole Cr(OH)₃)/(3 moles of KOH)

Let's calculate the moles KOH we have in 19 grams of KOH. 19g KOH/(56.1 g/mole KOH) = 0.3387 moles KOH

Now use the molar ratio to determine the moles Cr(OH)₃ produced:

We want the answer in grams Cr(OH)₃:

(0.3387 moles KOH)*(1 mole Cr(OH)₃)/(3 moles of KOH) = (0.1129 moles Cr(OH)₃)*(102.99 g/mole Cr(OH)₃) = 11.63 grams. Round to 12 grams for 2 sig figs.

Bob