Question

If 66 moles of NaOH combine with 26 moles of H,PO,, how many moles of H,O can form? What is the limiting reagent? 3 NAOH(aq)+H,PO,(aq) 3 H2O() + Na, PO, (aq) Step 1: Find the number of moles of H, O each starting material could form. 66 moles NaOH x moles H,O 26 moles H, PO x moles H,O Answer Bank 3 moles NaOH 1 mole H, PO 3 moles H,O I mole Na, PO Step 2: Determine the limiting reagent. Он, РО, NaOH

Answer 1

Available moles of NaOH = 66

Available moles of H3PO4 = 26

From balanced reaction;

1 mole H3PO4 requires 3 moles NaOH

So, 26 moles H3PO4 will require = 26*3 = 78 moles of NaOH

However, available moles of NaOH is 66 which is less than the required one i.e. 78.

Thus, NaOH is limiting reagent and H3PO4 is excess reagent.

Limiting reagent will drive the actual yield of product. Thus,

Again from reaction;

3 mole NaOH produces 3 mole H2O

So, 66 mole NaOH will produce = 3*66/3 = 66 moles of H2O

Hence, actual moles of H2O produced = 66

Again from reaction;

1 mole H3PO4 produces 3 mole H2O

So, 26 mole H3PO4 will produce = 3*26/1 = 78 moles of H2O

Excess moles of H3PO4 = 26 - (66/3) = 4 moles H3PO4

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