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How many grams of NaH2PO4 are needed to react with 36.29 mL of 0.250 M NaOH?

NaH2PO4(s) + 2 NaOH(aq) → Na3PO4(aq) + 2 H20(1)

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Final answer:

To find the grams of NaH₂PO₄needed, convert the volume of NaOH solution to moles using the molarity. Then, use the stoichiometry of the equation to determine the number of moles of NaH₂PO₄ needed. Finally, convert the moles of NaH₂PO₄ to grams using its molar mass.

Step-by-step explanation:

To determine the number of grams of NaH₂PO₄ needed to react with 36.29 mL of 0.250 M NaOH, we need to use the stoichiometry of the balanced equation.

The balanced equation is: NaH₂PO₄ (s) + 2 NaOH(aq) → Na₃PO4(aq) + 2 H₂O(1)

First, we convert the volume of NaOH solution to moles using the molarity:

Moles of NaOH = Volume (L) x Molarity = 36.29 mL x 0.250 mol/L = 0.00907 mol

Next, we use the stoichiometry of the equation to determine the number of moles of NaH₂PO₄ needed. From the balanced equation, we see that 1 mol of NaH2PO4 reacts with 2 mol of NaOH. Therefore, the number of moles of NaH₂PO₄ needed is twice the number of moles of NaOH:

Moles of NaH₂PO₄ needed = 2 x Moles of NaOH = 2 x 0.00907 mol = 0.0181 mol

Finally, we convert the moles of NaH₂PO₄ to grams using its molar mass:

Grams of NaH₂PO₄ = Moles x Molar mass = 0.0181 mol x 120.0 g/mol = 2.17 g

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