Given the following balanced equation:H3PO4(aq) + 3 KOH(aq) → K3PO4(aq) + 3 H2O(l)How many grams that potassium hydroxide are forced to totally neutralize 3.70 grams that phosphoric acid?
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3.70 g H3PO4 x (1 mole H3PO4/97.994g H3PO4) x (3mole KOH/1mole H3PO4) x (56.1056 g KOH/1 mole KOH) = 6.36 g KOH
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