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barium chloride plus sodium sulphate yields barium sulphate plus sodium chloride

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What is the balanced chemical equation for when barium chloride reacts with sodium sulfate?

This equation is BaCl2 (aq) + Na2SO4 (aq) -> 2 NaCl (aq) + BaSO4 (s).


What is the balanced equation of barium chloride and sodium sulfate include phases?

BaCl2(aq) + Na2SO4 (aq) ------> BaSO4(s) + 2NaCl(aq)


What is the reaction for Na2SO4 BaCl2?

Test for the anions,according to your word,i think you got the salts,so, use these reagents HCL,BACL2,Adding bacl2 and Hcl to Naso4-white ppt observed in soluble,but barium chloride will give no ppt...so probem solved


How do you balance BaCl2 and Na2SO4?

To balance the chemical equation between BaCl2 and Na2SO4, you need to ensure that the number of each type of atom is the same on both sides of the equation. The balanced equation is BaCl2 + Na2SO4 -> BaSO4 + 2NaCl.


What is the balanced equation for Barium Chloride plus Sodium Sulfate?

Ba2+ + [2Cl- + 2Na+] + SO42---> BaSO4 + [2Cl- + 2Na+]Ba2++ SO42- --> BaSO4


What volume of 0.131 M BaCl2 is required to react completely with 42.0 ml of .453 MNa2SO4. This is the net ionic equation for the reaction. Ba2 plus plus SO42---- BaSO4. Correct answer equals 145 ML?

first of all, you need to recognize that one mole of Na2SO4 is reacting with one mole of BaCL2. so find the moles of the NaSO4, then you automatically have the moles of the BaCL2. if you get the moles of the BaCL2, its easy to calculate the volume of it because you already have the MOLARITY. good luck


If 57.0 mL of BaCl2 solution is needed to precipitate all the sulfate ion in a 742 mg sample of Na2SO4 what is the molarity of the solution?

To find the molarity of the BaCl2 solution, first calculate the moles of Na2SO4 in the sample using its molar mass. Then, use the balanced chemical equation of the precipitation reaction to determine the moles of BaCl2 needed to react with the moles of Na2SO4. Finally, divide the moles of BaCl2 by the volume of the solution in liters (57.0 mL = 0.057 L) to find the molarity.


How many grams of solid barium sulfate form when 22.6 mL of 0.160 M barium chloride reacts with 54.6 mL of 0.055 M sodium sulfate?

Balanced equation first. BaCl2 + Na2SO4 -> 2NaCl + BaSO4 22.6 ml BaCl2 = 0.0226 liters 54.6 ml Na2SO4 = 0.0546 liters 0.160 M BaCl2 = moles BaCl2/0.0226 liters = 0.00362 moles BaCl2 0.055 M Na2SO4 = moles Na2SO4/0.0546 liters = 0.0030 moles Na2SO4 The ratio of BaCl2 to Na2SO4 is one to one, so either mole count wull drive this reaction. Use 0.0003 moles Na2SO4 0.0030 moles Na2SO4 (1 mole BaSO4/1 mole Na2SO4)(233.37 grams/1 mole BaSO4) = 0.700 grams of BaCO4 produced


What is the Chemical equation for the reaction of Sodium Sulphate with Barium Chloride?

The chemical equation for the reaction of sodium sulfate with barium chloride is: Na2SO4 + BaCl2 → 2NaCl + BaSO4. This is a double displacement reaction where the sodium and barium ions switch partners to form sodium chloride and barium sulfate.


What is the balanced equation for sodium sulfate plus barium chloride yields barium sulfate and sodium chloride?

Ooh. Somebody doesn't like doing their chem homework. Luckily I just happened to do that exact same equation. 1 BaCl2 +1 Na2S --> 2 NaCl(Salt! :D) +1 BaS


An aqueous solution BaCl2 is added to NaSO4 what reaction occurs?

When BaCl2 (barium chloride) is added to Na2SO4 (sodium sulfate), a precipitation reaction occurs, resulting in the formation of a white precipitate of barium sulfate (BaSO4). This is represented by the chemical equation: BaCl2 + Na2SO4 → BaSO4 + 2NaCl.


What is the name of the balanced equation of barium chloride and sodium sulfate?

Barium Chloride + Sodium Sulfate --> Barium Sulfate + Sodium Chloride BaCl2 + Na2So4 --> BaSO4 + 2NaCl It's called a Double Displacement reaction because Barium(Ba2+) and Sodium(Na+) displaces each other from their original anions. It's also called a Precipitation reaction because a white precipitate is formed after the reaction due to Barium Sulfate(BaSO4) as it is insoluble.