Stoichiometry help please??!!?

Question

How many grams of NaH2PO4 are needed to react with 38.74 mL of a 0.275 M NaOH?

NaH2PO4 + 2NaOH ---> Na3PO4 + 2H2O

Could you not only give the answer, but please show how to do it? I think these are the steps...

1. Balance the equation.

2. Find the number of moles of one substance.

3. relate it to the stoichiometrically equivalent number of moles of another substance.

4. convert to desired units.

Answer 1

Atomic weights: Na=23 H=1  P=31  O=16  NaH2PO4=121

Let the NaOH solution be called S.

38.74mLS x 0.275molNaOH/1000mLS x 1molNaH2PO4/2molNaOH x 121gNaH2PO4/1molNaH2PO4 = 0.6445g NaH2PO4 to four significant figures

Answer 2

You are right on!

1. The equation is already balanced.

2. Molarity times volume in L will give you the number of moles of NaOH solution.

3. The equation shows that 1mol of NaH2PO4 completely neutralizes 2 moles of NaOH.

4. From #3 you can figure out how many moles of NaH2PO4 there should be (since you  calculated number of moles of NaOH in #2). Now use the molar mass of Na2HPO4 to calculate the mass from its number of moles (hint: number of moles times its molar mass will give you the mass).