Given the following balanced equation: H3PO4(aq) + 3 KOH(aq) → K3PO4(aq) + 3 H2O(l) How many grams of potassium hydroxide are required to completely neu

Question

Given the following balanced equation:

H3PO4(aq) + 3 KOH(aq) → K3PO4(aq) + 3 H2O(l)

How many grams of potassium hydroxide are required to completely neutralize 3.70 grams of phosphoric acid?

Answer 1

H3PO4(aq) + 3 KOH(aq) → K3PO4(aq) + 3 H2O(l)

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

 

Better check my math...