Explain in word how you would prepae 250mL of a 3.000 M solution of Ba(OH)2 starting form barium hydroxide and water.....
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....
chemistry - DrBob222, Sunday, October 10, 2010 at 5:14pm
The definition of molarity is
M = moles/L of solution.
The secret to the getting the problem right is in how you say it.If you say "add 0.300 mole Ba(OH)2 (which is 171 grams) to 1 L of water" you will get no credit for the problem. If you say "add 171 grams Ba(OH)2 to a 1 L flask, add some water, swirl to dissolve all of the solid, then make to a final volume of 1 L, you will get credit. The difference in the two statements, if you look closely, is that the first will be a solution with MORE THAN 1 L of solution because the Ba(OH)2 occupies some volume so the final volume will not be 1 L. I hope you see the difference.
chemistry - pete, Sunday, October 10, 2010 at 5:29pm
thank you but how is 171 moles/L equals to 250 mL of the 3.000 M solutions of Ba(OH)2
chemistry - DrBob222, Sunday, October 10, 2010 at 9:34pm
It isn't. Thank you for the follow up. The 171 g Ba(OH)2 prepares 1 L of 3.00 M solution. To prepare just 250 mL one takes 1/4 of the 171 g.
I became so interested in the fine details of how the solution was prepared that I completely forgot we weren't making a L.