Explain in word how you would prepae 250mL of a 3.000 M solution of Ba(OH)2 starting form barium hydroxide and water.....
2 equation
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....
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.
thank you but how is 171 moles/L equals to 250 mL of the 3.000 M solutions of Ba(OH)2
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.
To prepare a 250mL solution of 3.000 M Ba(OH)2, you would need to follow these steps:
1. Determine the amount of Ba(OH)2 needed. Molarity is defined as moles of solute per liter of solution. In this case, the molarity is 3.000 M and the volume of the solution is 250mL, which is equal to 0.25L. So, the number of moles of Ba(OH)2 needed can be calculated as follows:
moles of Ba(OH)2 = molarity * volume of solution
moles of Ba(OH)2 = 3.000 M * 0.25 L
2. Convert the moles of Ba(OH)2 to grams. To do this, you need to know the molar mass of Ba(OH)2, which can be calculated by adding up the atomic masses of each element:
molar mass of Ba(OH)2 = atomic mass of Ba + (2 * atomic mass of O) + (2 * atomic mass of H)
molar mass of Ba(OH)2 = (137.33 g/mol) + (2 * 16.00 g/mol) + (2 * 1.01 g/mol)
3. Calculate the grams of Ba(OH)2 needed by multiplying the moles of Ba(OH)2 by its molar mass:
grams of Ba(OH)2 = moles of Ba(OH)2 * molar mass of Ba(OH)2
To find the mass of cobalt (III) sulfide formed when 1318.89 grams of cobalt (III) iodide and 551.35 grams of potassium sulfide are reacted, you would follow these steps:
1. Determine the number of moles of cobalt (III) iodide and potassium sulfide using their respective molar masses (which can be calculated similar to the molar mass of Ba(OH)2).
2. Use the balanced equation to determine the stoichiometry between cobalt (III) iodide and cobalt (III) sulfide. From the equation, you can see that 2 moles of CoI3 react to form 1 mole of Co2S3. This means that the mole ratio is 2:1.
3. Calculate the moles of cobalt (III) sulfide formed using the mole ratio determined from the balanced equation.
4. Finally, calculate the mass of cobalt (III) sulfide formed by multiplying the moles of Co2S3 by its molar mass (calculated similarly to the molar mass of Ba(OH)2).
Using the given mass of the potassium sulfide is not necessary in this calculation as only the mass of cobalt (III) iodide is provided.