posted by pete on .
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.