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 the answer to be 321 grams of Co2S3>>> and was i suppose to use the given mass of the Potassium sulfide as well....?
See my response to this problem above.
Well, I have to say, it sounds like you did your chemistry homework! Great job!
To calculate the mass of cobalt (III) sulfide formed, you need to use the concept of stoichiometry, which relates the number of moles of one substance to another in a chemical equation.
First, you need to convert the given masses of cobalt (III) iodide and potassium sulfide into moles using their respective molar masses. Cobalt (III) iodide (CoI3) has a molar mass of 416.69 g/mol, and potassium sulfide (K2S) has a molar mass of 110.26 g/mol.
1318.89 g of CoI3 = (1318.89 g / 416.69 g/mol) = 3.16 moles of CoI3
551.35 g of K2S = (551.35 g / 110.26 g/mol) = 5 moles of K2S
From the balanced equation, we can see that the stoichiometric ratio of CoI3 to Co2S3 is 2:1, meaning that for every 2 moles of CoI3, 1 mole of Co2S3 is formed.
Therefore, if we have 3.16 moles of CoI3, we can expect to form (3.16 moles / 2) = 1.58 moles of Co2S3.
Finally, we can convert moles of Co2S3 into grams by multiplying by its molar mass. The molar mass of Co2S3 is 223.86 g/mol.
1.58 moles of Co2S3 = (1.58 moles * 223.86 g/mol) = 353.50 grams
So, according to my calculations, you should expect to form approximately 353.50 grams of cobalt (III) sulfide. Keep up the good work!
To find the mass of cobalt (III) sulfide formed, you need to follow these steps:
Step 1: Calculate the number of moles for each given substance using their molar masses.
- Molar mass of CoI3 = (58.9332 g/mol) + 3*(126.9045 g/mol) = 391.8347 g/mol
- Moles of CoI3 = 1318.89 g / 391.8347 g/mol = 3.3650 mol
- Molar mass of K2S = (39.0983 g/mol) + 2*(32.066 g/mol) = 110.2583 g/mol
- Moles of K2S = 551.35 g / 110.2583 g/mol = 5.0009 mol
Step 2: Determine the limiting reactant.
- The balanced equation tells us that the stoichiometric ratio between CoI3 and K2S is 2:3.
- The ratio of moles is 3.3650 mol CoI3 to 5.0009 mol K2S, which simplifies to 0.6739:1.
- Since the ratio is less than 1, CoI3 is the limiting reactant.
Step 3: Determine the number of moles of cobalt (III) sulfide formed.
- From the balanced equation, we know that for every 2 moles of CoI3, 1 mole of Co2S3 is formed.
- Moles of Co2S3 = (0.6739 mol CoI3) / (2 mol CoI3/1 mol Co2S3) = 0.3369 mol
Step 4: Convert the number of moles to grams of cobalt (III) sulfide.
- Molar mass of Co2S3 = (58.9332 g/mol) + 3*(32.065 g/mol) = 165.9602 g/mol
- Mass of Co2S3 = 0.3369 mol * 165.9602 g/mol = 55.9008 g
So, the correct answer is 55.9008 grams of cobalt (III) sulfide will form.
To determine how many grams of cobalt (III) sulfide will form, we need to use the concept of stoichiometry.
First, let's calculate the number of moles of cobalt (III) iodide and potassium sulfide given the masses provided.
Molar mass of cobalt (III) iodide (CoI3) = atomic mass of Co + 3 * atomic mass of I = 58.9332 g/mol + 3 * 126.9045 g/mol = 434.6757 g/mol
Molar mass of potassium sulfide (K2S) = 2 * atomic mass of K + atomic mass of S = 2 * 39.0983 g/mol + 32.06 g/mol = 110.2526 g/mol
Moles of cobalt (III) iodide = mass of cobalt (III) iodide / molar mass of cobalt (III) iodide = 1318.89 g / 434.6757 g/mol ≈ 3.034 mol
Moles of potassium sulfide = mass of potassium sulfide / molar mass of potassium sulfide = 551.35 g / 110.2526 g/mol ≈ 5.000 mol
Now, let's check the stoichiometric ratio between the reactants and the product.
From the balanced equation: 2CoI3 + 3K2S = Co2S3 + 6KI
We see that the ratio of cobalt (III) iodide to cobalt (III) sulfide is 2:1.
Since we have 3.034 moles of cobalt (III) iodide, we can determine the moles of cobalt (III) sulfide by dividing it by 2:
Moles of cobalt (III) sulfide = Moles of cobalt (III) iodide / 2 = 3.034 mol / 2 = 1.517 mol
Finally, we can calculate the mass of cobalt (III) sulfide using its molar mass.
Molar mass of cobalt (III) sulfide (Co2S3) = 2 * atomic mass of Co + 3 * atomic mass of S = 2 * 58.9332 g/mol + 3 * 32.06 g/mol = 324.9498 g/mol
Mass of cobalt (III) sulfide = Moles of cobalt (III) sulfide * molar mass of cobalt (III) sulfide = 1.517 mol * 324.9498 g/mol ≈ 492.3926 g
Therefore, the mass of cobalt (III) sulfide formed when 1318.89 grams of cobalt (III) iodide are combined with 551.35 grams of potassium sulfide is approximately 492.3926 grams. The answer you found (321 grams) seems to be incorrect based on these calculations.
Yes, you needed to use the given mass of potassium sulfide in order to determine the amount of cobalt (III) sulfide formed. The stoichiometry of the reaction is crucial for these calculations.