Question 8 (Mandatory) (10 points) When aqueous solutions of cobalt (III) bromid

Question

Question 8 (Mandatory) (10 points) When aqueous solutions of cobalt (III) bromide and potassium sulfide are allowed to mix, a precipitate of cobalt (III) sulfide and aqueous potassium bromide are formed. 1. Write a balanced chemical equation for the reaction. 2. Write the net ionic equation. 3. How many grams of cobalt (III) bromide are required to react 17 g potassium sulfide? 4. How many grams of cobalt (III) sulfide are formed from the reaction of 45 g cobalt (III) bromide in the presence of excess potassium sulfide? 5 Hide hint for Question 8 When drawing the arrow, do the following -> Use only 1 space between each input...ex. 675 g or 2 NaC H2S03- >, NaCI (s) Save

Explanation / Answer

1. Balanced chemical equation for the reaction is

2 CoBr3 (aq) + 3 K2S (aq) --> Co2S3 (s) + 6 KBr (aq)

2. The net ionic equation is

2 Co3+(aq) + 3 S2- (aq) --> Co2S3 (s)  

3. given mass of   K2S is 17g,

17 g of K2S = (17g/110.262 g/mol) =0.1541 mol

from the balanced equation we can see that 3 mol of K2S reacts with 2 mol of  CoBr3

hence mol of CoBr3 required = (2/3)x0.1541 mol = 0.1027 mol

therefore mass of CoBr3 required = 0.1027 mol x 298.6452 g/mol =30.696 g

Hence 30.696 gram of Cobalt(III) Bromide is required.

4. 45 gram of Cobalt(III) Bromide is present

45 gram of Cobalt(III) Bromide = (45g/298.6452g/mol) = 0.1506 mol

from the balanced equation we can see 2 mol of Cobalt(III) Bromide forms 1 mol of cobalt(III) Sulphide

hence , mol of  cobalt(III) Sulphide formed = (1/2) x 0.1506 mol =0.0753 mol

therefore the mass of cobalt(III) Sulphide formed =0.0753 molx 214.0614 g/mol = 16.118 g.

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