If 0.2500 grams of barium of potassium sulfate react with an sulfate will be pro

Question

If 0.2500 grams of barium of potassium sulfate react with an sulfate will be produced? What two things are done to maximize the size of the BaS0_4 crystals? In this experiment how are the two products separated after the reaction is completed? Describe where each product is located at the end of the separation. Write the net ionic equation for the reaction of sodium sulfate with barium iodide. A student reacts 0.3000 grams of a sulfate-containing unknown with excess barium chloride. If 0.6500 grams of barium sulfate is produced, what is the % sulfate in the unknown?

Explanation / Answer

1) K2SO4 + BaCl2 ------> 2KCl + BaSO4

Mass of Pottasium sulfate taken = 0.25

moles of K2SO4 = 0.25/174 = 0.00144 = moles of BaSO4 produced

Mass of BaSO4 produced = 233 x 0.00144 = 0.3352 gm of barium sulfate was produced.

2)

To maximize the size of crystals

---> Solution must be slightly acidic in nature.

----> Concentration of Sulphate ions must be limited otherwise higher concentration leads to occlusion of barium sulphate crystals

4) Na2SO4 + BaI -----> 2 NaI + BaSO4

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