"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.
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?
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