Sulfuric acid reacts with NaOH: (a) Write a balanced equation for the reaction producing \(\mathrm{Na}_{2} \mathrm{SO}_{4}\). (b) How many milliliters of \(0.10 \mathrm{M} \mathrm{NaOH}\) are needed to react with \(0.0050 \mathrm{~mol} \mathrm{H}_{2} \mathrm{SO}_{4}\) ? (c) How many grams of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) will also form?

In any chemical reaction, a balanced chemical equation is crucial. It ensures that the law of conservation of mass is upheld—meaning the same number of atoms for each element is present on both sides of the equation. For the reaction between sulfuric acid \( \mathrm{H}_2\mathrm{SO}_4 \) and sodium hydroxide \( \mathrm{NaOH} \), the balanced equation is:
\[ \mathrm{H}_2\mathrm{SO}_4 + 2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + 2 \mathrm{H}_2\mathrm{O} \]
This equation shows that one molecule of sulfuric acid reacts with two molecules of sodium hydroxide to produce one molecule of sodium sulfate \( \mathrm{Na}_2\mathrm{SO}_4 \) and two molecules of water \( \mathrm{H}_2\mathrm{O} \). This understanding helps us figure out quantities of reactants and products.

The mole ratio in a balanced chemical equation tells us the proportions of reactants and products involved. From our balanced equation for \( \mathrm{H}_2\mathrm{SO}_4 \) and \( \mathrm{NaOH} \), we see the ratio is:
\[ \mathrm{H}_2\mathrm{SO}_4 : \mathrm{NaOH} = 1:2 \]
This means that one mole of sulfuric acid requires two moles of sodium hydroxide to react completely. This ratio is essential for calculations because it allows us to convert between moles of different substances in the reaction. For example, knowing you have 0.0050 moles of sulfuric acid, you need twice that amount in moles of sodium hydroxide, which is 0.010 moles.

Molarity (M) is a way of expressing the concentration of a solution. It is defined as:
\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]
To find the volume of a \( \mathrm{NaOH} \) solution needed to react with our sulfuric acid, we use the molarity equation rearranged:
\[ \text{Volume (L)} = \frac{\text{moles}}{M} \]
Substituting in our values (0.010 moles \( \mathrm{NaOH} \) and 0.10 M solution):
\[ \text{Volume (L)} = \frac{0.010 \text{ moles}}{0.10 \text{ M}} = 0.10 \text{ L} \]
Converting liters to milliliters (since 1 L = 1000 mL):
\[ 0.10 \text{ L} = 100 \text{ mL} \]
So, 100 mL of 0.10 M \( \mathrm{NaOH} \) is needed.

After determining the moles of product, you can find its mass. The molar mass (molecular weight) of a substance is the mass of one mole of that substance. For sodium sulfate \( \mathrm{Na}_2\mathrm{SO}_4 \), the molar mass is calculated by adding the atomic masses of all atoms in the formula:
\[ 2\mathrm{Na} + \mathrm{S} + 4\mathrm{O} = 2(22.99) + 32.06 + 4(16.00) = 142.04 \text{ g/mol} \]
Knowing we produced 0.0050 moles of \( \mathrm{Na}_2\mathrm{SO}_4 \), the mass is:
\[ \text{Mass} = 0.0050 \text{ moles} \times 142.04 \text{ g/mol} = 0.7102 \text{ g} \]
Therefore, 0.7102 grams of sodium sulfate is formed.