How many grams of \(\mathrm{NaH}_{2} \mathrm{PO}_{4}\) are needed to react with \(43.74 \mathrm{~mL}\) of \(0.285 \mathrm{M} \mathrm{NaOH} ?\) $$ \mathrm{NaH}_{2} \mathrm{PO}_{4}(s)+2 \mathrm{NaOH}(a q) \longrightarrow \mathrm{Na}_{3} \mathrm{PO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(I) $$

Calculating the number of moles is a fundamental concept in stoichiometry. It tells us how many particles (atoms, molecules, or ions) are in a given amount of substance. Here's how you do it:

If you know the molarity (M) of a solution and its volume in liters (L), you can find the number of moles. The formula is straightforward:

• \( \text{moles} = \text{Molarity} \times \text{Volume} \)

For example, in the provided exercise, we start by calculating the moles of \(\text{NaOH}\) given its molarity (0.285 M) and volume (43.74 mL, which is 0.04374 L after converting from mL to L). Plugging these values into the formula gives us:

• \( \text{moles of NaOH} = 0.285 \times 0.04374 = 0.01246 \text{ moles} \)

Knowing the number of moles allows us to proceed with other calculations, such as determining the amount of reactant or product involved in a chemical reaction.

A balanced chemical equation is crucial for solving stoichiometry problems as it ensures the conservation of mass and atoms. In our example, the balanced equation is:
\[ \text{NaH}_{2} \text{PO}_{4}(s) + 2 \text{NaOH}(aq) \rightarrow \text{Na}_{3} \text{PO}_{4}(aq) + 2 \text{H}_{2} \text{O}(l) \]
This equation tells us:

• 1 mole of \(\text{NaH}_{2} \text{PO}_{4}\) reacts with 2 moles of \(\text{NaOH}\).
• This produces 1 mole of \(\text{Na}_{3} \text{PO}_{4}\) and 2 moles of \(\text{H}_{2} \text{O}\).

We use this stoichiometric relationship to find the moles of \(\text{NaH}_{2} \text{PO}_{4}\) required to react with the moles of \(\text{NaOH}\) calculated earlier. Because it takes 2 moles of \(\text{NaOH}\) to react with 1 mole of \(\text{NaH}_{2} \text{PO}_{4}\), we use the ratio:

• \( \text{moles of NaH}_{2} \text{PO}_{4} = \frac{\text{moles of NaOH}}{2} = \frac{0.01246}{2} = 0.00623 \text{ moles} \)

Having this balanced equation makes sure we adhere to the law of conservation of mass, which is fundamental in chemical reactions.

Converting moles to grams involves using the molar mass of a substance. The molar mass is the mass of one mole of a given substance (usually in g/mol). It can be calculated by summing the atomic masses of all atoms in the molecule. For \(\text{NaH}_{2} \text{PO}_{4}\), you sum the atomic masses of sodium (Na), hydrogen (H), phosphorus (P), and oxygen (O):

• \( \text{Molar mass of NaH}_{2} \text{PO}_{4} = 23 + 2 \times 1 + 31 + 4 \times 16 = 120 \text{ g/mol} \)

Once the molar mass is known, you can convert moles to grams using the formula:

• \( \text{mass (g)} = \text{moles} \times \text{molar mass} \)

Plugging in our values, we get:

• \( \text{mass} = 0.00623 \text{ moles} \times 120 \text{ g/mol} = 0.7476 \text{ g} \)

This final conversion tells us we need 0.7476 grams of \(\text{NaH}_{2} \text{PO}_{4}\) to react with 43.74 mL of 0.285 M \(\text{NaOH}\). Knowing how to perform these conversions and calculations is essential for solving stoichiometry problems accurately.