After a lab experiement, you need to neutralize 13.22 grams of barium hydroxide

Question

After a lab experiement, you need to neutralize 13.22 grams of barium hydroxide dissolved in water. You decide to do this by adding some solid phosphoric acid (H3PO4), resulting in a precipitate of barium phosphate.

A) Write a balanced chemical equation fo this process.

B) If you add 5.95g of H3PO4, is that enough to neutralize the barium hydroxide. Calculate and explain.

C) What is the minimum number of grams of phosphoric acid you need to neutralize all of the barium hydroxide?

D) How many grams of barium phosphate are produced?

Explanation / Answer

The balanced reaction is

3Ba(OH)2 + 2H3PO4 --------->Ba3(PO4)2 +3 H2O

3 moles of Ba(OH)2 reacts with 2 moles of H3PO4 to give 1 mole of Ba3(PO4)2

theoretical molar ratio of Ba(OH)2: H3PO4= 3:2 or 1.5:1

moles= mass/molar mass, molar mass of Ba(OH)2= 171 g/mole and H3PO4= 98 g/mole

moles of Ba(OH)2= 13.22/171 =0.077 and moles of H3PO4= 5.95/98 =0.061

actual molar ratio of Ba(OH)2: H3PO4= 0.077 :0.061 =1.26 :1

Excess reagent is H3PO4 and limiting reagent is Ba(OH)2. so added, H3PO4 completely neutralizes Ba(OH)2 and some amounrt of H3PO4 is left unreacted. minimum amount required per 0.077 moles of Ba(OH)2= 0.077*2/3=0.052 moles or 0.052*98 gm =5.1 gm

moles of Ba3(PO4)2 formed = 0.077/3=0.026 , molar mass of Ba3(PO4)2= 465

mass of Ba3(PO4)2 formed= 0.026*465 gm =12.09 gm

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