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Given 5.0 moles of KOH and 2.0 moles of H3PO4, how many moles of K3PO4 can be prepared?which is the limiting reagent? which is the excess? ?
Hii pls help i dont understand
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- Roger the MoleLv 77 months ago
3 KOH + H3PO4 → K3PO4 + 3 H2O
5.0 moles of KOH would react completely with 5.0 x (1/3) = 1.7 moles of H3PO4, but there is more H3PO4 present than that, so H3PO4 is in excess and KOH is the limiting reagent.
(5.0 mol KOH) x (1 mol K3PO4 / 3 mol KOH) = 1.7 mol K3PO4
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