consider the following balanced chemical reaction: 2 Col3+3 K2S-> Co2S3+6 Kl
assuming all that can react will react, how many grams of cobalt (III) sulfate will form when 1318.89 grams of cobalt (III) iodide are combined with 551.35 grams of potassium sulfide?
"how many grams of cobalt (III) sulfate will form when 1318.89 grams of cobalt (III) iodide are combined with 551.35 grams of potassium sulfide? "
Taking the question literally, 0 grams of cobalt (III) sulfate are formed as the product is cobalt (III) sulfide.
Assuming that this is a typo then you need to calculate how many moles of each of the two starting materials are present using
number of moles = mass/molar mass
from this you can then determine which of the two is the limiting reagent.
Hence calculate the number of moles of Co2S3 formed, from which you can calculate the mass.
To determine the number of grams of cobalt (III) sulfate that will form, we need to use stoichiometry - the numerical relationship between reactants and products in a balanced chemical equation.
First, let's find the limiting reactant, which is the reactant that is completely consumed and determines the maximum amount of product formed.
1. Convert the given masses of cobalt (III) iodide (CoI3) and potassium sulfide (K2S) to moles. To do this, divide the given masses by their respective molar masses.
Molar mass of CoI3: Co (Atomic Mass: 58.93 g/mol) + I (Atomic Mass: 126.9 g/mol) x 3 = 442.63 g/mol
Number of moles of CoI3 = 1318.89 g / 442.63 g/mol
Molar mass of K2S: K (Atomic Mass: 39.10 g/mol) x 2 + S (Atomic Mass: 32.07 g/mol) = 110.27 g/mol
Number of moles of K2S = 551.35 g / 110.27 g/mol
2. Determine the stoichiometric ratio of CoI3 to Co2S3 by comparing the coefficients in the balanced chemical equation:
2 CoI3 + 3 K2S -> Co2S3 + 6 Kl
The ratio of CoI3 to Co2S3 is 2:1.
3. Determine the limiting reactant by comparing the moles of CoI3 and K2S. The reactant that produces fewer moles of product is the limiting reactant.
Moles of Co2S3 produced from CoI3 = (Number of moles of CoI3) / 2
Moles of Co2S3 produced from K2S = (Number of moles of K2S) * 1
Compare the two values to find the limiting reactant.
4. Once you determine the limiting reactant, calculate the number of moles of Co2S3 produced using the stoichiometric ratio from the balanced chemical equation.
5. Finally, convert the moles of Co2S3 to grams using the molar mass of Co2S3 (280.95 g/mol):
Mass of Co2S3 = (Number of moles of Co2S3) x (Molar mass of Co2S3)
By following these steps, you can calculate the grams of cobalt (III) sulfate formed when the given reactants are combined.