2 equation
2CoI3 + 3 K2S= Co2S3 + 6KI
2 moles of Cobalt (III) iodide + 3 moles of Potassium sulfide = cobalt (III) sulfide + 6moles of Potassium Iodide
assuming all that can react will react, how many grams of cobalt (iii) sulfide will form when 1318.89 grams of cobalt (III) iodide are combined with 551.35 grams of potassuim sulfide..... <<<< I found it to be 321 grams of Co2S3>>> and was suppose to use the given mass of the Potassium sulfide as well....
You got lucky. CoI3 is the limiting reagent; therefore, the mass Co2S3 depends upon the starting amount of CoI3 and 321.09 g is the correct answer. (You are allowed more places than 321).
Here is how you figure in the K2S.
moles CoI3 1318.89/439.6466 = rounds to 3.0 moles CoI3.
moles K2S = 551.35/110.26 = rounds to 5.0
Now convert these moles to moles of the product.
3.0 moles CoI3 x (1 mole Co2S3/2 moles CoI3) = 3.0 x (1/2) = 1.5 moles Co2S3.
5.0 moles K2S x (1 mole Co2S3/3 moles CoI3) = 5.0 x (1/3) = 1.667
Both answers can't be right; one of them must be wrong. The correct answer in limiting reagent problem is ALWAYS the smaller value and the reagent producing that value is the limiting reagent. Therefore, CoI3 is the limiting reagent.
Now 1.5 moles Co2S3 x molar mass Co2S3 = xx g.
How do you know when you have a limiting reagent problem? Easy. When you have BOTH reagents given instead of just one.
To find the mass of cobalt (III) sulfide formed, we first need to calculate the number of moles of cobalt (III) iodide and potassium sulfide using their respective masses.
1. Start by calculating the number of moles of Cobalt (III) iodide (CoI3):
- Given mass of CoI3 = 1318.89 grams
- Molar mass of CoI3 = 286.70 grams/mole
- Moles of CoI3 = Given mass / Molar mass
= 1318.89 / 286.70
≈ 4.596 moles
2. Next, calculate the number of moles of Potassium sulfide (K2S):
- Given mass of K2S = 551.35 grams
- Molar mass of K2S = 110.26 grams/mole
- Moles of K2S = Given mass / Molar mass
= 551.35 / 110.26
≈ 5 moles
3. Now, let's determine the limiting reagent. The reagent that yields the smallest number of moles is the limiting reagent.
- From the balanced equation, 2 moles of CoI3 react with 3 moles of K2S to form 1 mole of Co2S3.
- Based on the ratio, we need 2/3 × moles of K2S to react with moles of CoI3.
- Moles of K2S required = (2/3) × moles of CoI3
= (2/3) × 4.596
≈ 3.06 moles
Here, we can see that the moles of K2S (5 moles) are in excess compared to the required moles (3.06 moles). Therefore, CoI3 acts as the limiting reagent.
4. Calculate the moles of Co2S3 formed based on the limiting reagent:
- From the balanced equation, 2 moles of CoI3 react to form 1 mole of Co2S3.
- Moles of Co2S3 = 1/2 × moles of CoI3
= 1/2 × 4.596
≈ 2.298 moles
5. Finally, calculate the mass of Co2S3 formed:
- Given that Molar mass of Co2S3 = 243.92 grams/mole
- Mass of Co2S3 = Moles of Co2S3 × Molar mass
= 2.298 × 243.92
≈ 560 grams
Therefore, when 1318.89 grams of Cobalt (III) iodide reacts with 551.35 grams of Potassium sulfide, approximately 560 grams of cobalt (III) sulfide will be formed.