Question

How many grams of NaH2PO4 are needed to react with 32.94 mL of 0.280 M NaOH? NaH2PO4(s) + 2 NaOH(aq) Na3PO4(aq) + 2 H2O(l)

Answer 1

from the equation NaH2PO4(s) + 2 NaOH(aq) Na3PO4(aq) + 2 H2O(l)

0.280 M NaOH solution means 0.28 moles of NaoH is dissolved in 1L of water. = 0.28 moles in 1000 mL.

So 0.00028 moles in 1mL. 0.00028*32.94 = 0.0092 moles in 32.94 mL.

From the reaction formula, 1 mol NaH2PO4 reacts with 2 moles of NaOH and produces 1 mole of Na3PO4 and 2 moles of H2O

so 0.0092 moles NaOH needs 0.0092/2 (half of it) moles of NaH2PO4 = 0.0046 moles

(NaH2PO4 Molar mass: 119.98 g/mol)

0.0046moles NaH2PO4 = 0.0046*119.98 g = 0.55 g