What mass of precipitate (in g) is formed when 250.0 mL of 0.200 M CoCl₂ is mixed with excess KOH in the following chemical reaction? CoCl₂(aq) + 2 KOH(aq) → Co(OH)₂(s) + 2 KCl(aq)

Answer 1

Hi Hattie!

To determine the mass of precipitant (or solid) Co(OH)₂ formed in this reaction, you will first need to calculate the number of moles of CoCl₂ present in the reaction. To do so, you will need to convert the mL of the CoCl₂ solution into liters (L):

250.0 mL x ( 1L/1000 mL) = 0.250 mL.

Next, remember that units of molarity are really moles/liter. Thus, you can multiply the volume in L you just calculated by the molarity given in order to get moles of CoCl₂:

0.250 L x (0.200 moles CoCl₂ / 1 L) = 0.0500 moles CoCl₂

Now you can use the balanced reaction to convert moles of CoCl₂ to moles of Co(OH)₂ by realizing that there is one mole of Co(OH)₂ formed for every one mole of CoCl₂ consumed:

0.0500 moles CoCl₂ x (1 mole Co(OH)₂ / 1 mole CoCl₂) = 0.0500 moles Co(OH)₂

Lastly, you can convert the moles of Co(OH)₂ to grams of Co(OH)₂ using the molar mass of Co(OH)₂ (92.948 grams/mole):

0.0500 moles Co(OH)₂ x (92.948 grams Co(OH)₂/ 1 mole Co(OH)₂) = 4.65 grams Co(OH)₂

Hope this helps!