If 0.5 mole of calcium bromine is mixed with 0.2 Mole of potassium phosphate calculate the number of moles of calcium phosphate that can be formed
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Balanced equation for reaction between BaCl2 and Na3PO4 is as follows:
3BaCl2 + 2Na3PO4 → Ba3(PO4)2 + 6NaCl
3 moles of BaCl2 react with 2 moles of Na3PO4 to give 1 mole of Ba3(PO4)2
0.5 moles of BaCl2 will react with (2/3) x 0.5 = 0.33 moles of Na3PO4
Available moles of Na3PO4 = 0.2
Hope this will help you..... ✌
So, Na3PO4 is the limiting reagent
Now, 2 moles of Na3PO4 give 1 mole of Ba3(PO4)2
So, 0.2 moles of Na3PO4 will give 1/2) x 0.2 = 0.1 mole of Ba3(PO4)2
Hence, maximum number of moles of Ba3(PO4)2 formed = 0.1
3BaCl2 + 2Na3PO4 → Ba3(PO4)2 + 6NaCl
3 moles of BaCl2 react with 2 moles of Na3PO4 to give 1 mole of Ba3(PO4)2
0.5 moles of BaCl2 will react with (2/3) x 0.5 = 0.33 moles of Na3PO4
Available moles of Na3PO4 = 0.2
Hope this will help you..... ✌
So, Na3PO4 is the limiting reagent
Now, 2 moles of Na3PO4 give 1 mole of Ba3(PO4)2
So, 0.2 moles of Na3PO4 will give 1/2) x 0.2 = 0.1 mole of Ba3(PO4)2
Hence, maximum number of moles of Ba3(PO4)2 formed = 0.1
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