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If 0.5 mol of BaCl2 is mixed with 0.20 mol of Na3PO4, the maximum amount of Ba3(PO4)2 that can be formed is
3 BaCl2 + 2 Na3PO4 = Ba3(PO4)2 + 6NaCl
Here clearly BaCl2 is the limiting agent
So 3 moles of bacl2 gives 1 mole of ba3(po4)2
so 0.5 mole will give 0.2 mole
alloy formation
90ml of an unknown concentrated chlorine was diluted with 2250l of water.The dilution now is 0.08mg/l. what is the initial concentration of the chlorine before dilution?
reaction of mono hydric alcohol
s block
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