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 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.