Question

Calculate the volume, in milliliters, of \(0.125-\mathrm{M} \mathrm{HNO}_{3}\) required to react completely with \(1.30 \mathrm{~g} \mathrm{Ba}(\mathrm{OH})_{2}\). \(2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{~s}) \longrightarrow \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell)\)

Short answer

121.44 mL of 0.125-M HNO3 is needed.

Step-by-step solution

Write the Balanced Chemical Equation

The balanced chemical equation for the reaction is already provided in the problem: \[2 \mathrm{HNO}_{3}(aq) + \mathrm{Ba(OH)}_{2}(s) \longrightarrow \mathrm{Ba(NO}_{3})_{2}(aq) + 2 \mathrm{H}_{2} \mathrm{O}(\ell)\] This equation shows that 2 moles of \(\mathrm{HNO}_{3}\) react with 1 mole of \(\mathrm{Ba(OH)}_{2}\).

Calculate Moles of \(\mathrm{Ba(OH)}_{2}\)

Using the molar mass of \(\mathrm{Ba(OH)}_{2}\), which is approximately \(171.34\) g/mol, calculate the number of moles in \(1.30\) grams: \[\text{Moles of } \mathrm{Ba(OH)}_{2} = \frac{1.30 \, \text{g}}{171.34 \, \text{g/mol}} \approx 0.00759 \, \text{mol}\]

Determine Moles of \(\mathrm{HNO}_{3}\) Needed

According to the stoichiometry of the reaction, it takes 2 moles of \(\mathrm{HNO}_{3}\) to react with 1 mole of \(\mathrm{Ba(OH)}_{2}\). Therefore, the moles of \(\mathrm{HNO}_{3}\) needed are: \[0.00759 \, \text{mol} \times 2 = 0.01518 \, \text{mol of } \mathrm{HNO}_{3}\]

Calculate Volume of \(\mathrm{HNO}_{3}\) Solution

To find the volume of the \(\mathrm{HNO}_{3}\) solution required, use the molarity formula \(M = \frac{\text{moles}}{\text{liters}}\): \[0.125 = \frac{0.01518}{V} \] Solve for \(V\): \[V = \frac{0.01518}{0.125} \approx 0.12144 \, \text{L}\] Convert this to milliliters: \[0.12144 \, \text{L} \times 1000 \, \text{mL/L} = 121.44 \, \text{mL}\]

Key concepts

Molarity Calculation

Molarity is a way to express the concentration of a solution. It tells you how many moles of a substance are dissolved in one liter of solution. The formula to calculate molarity (M) is \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \). To find the volume of a solution based on its molarity and the number of moles, we rearrange the formula to solve for liters: \( V = \frac{\text{moles}}{M} \). If you know how many moles of an acid or base you need for a reaction, and the concentration of your solution, this formula helps you find out exactly how much solution to use.
In the provided exercise, we calculated the volume of the \(0.125-\mathrm{M} \mathrm{HNO}_{3}\) required. First, we determined the moles of \(\mathrm{HNO}_{3}\) necessary using stoichiometry. Then, using the molarity formula, we solved for volume \( V \) and found it to be approximately \(0.12144\) liters, which is equivalent to \(121.44\) milliliters.

Balanced Chemical Equations

Balanced chemical equations are essential for understanding how reactions occur. They show the proportions in which reactants combine and the products they form. Each side of the equation must have the same number of each type of atom. This ensures the law of conservation of mass is maintained — matter is not created or destroyed.
A balanced equation is crucial in stoichiometry, which uses these equations to determine the relationships between reactants and products. In our exercise, the equation \(2 \mathrm{HNO}_{3}(aq) + \mathrm{Ba(OH)}_{2}(s) \longrightarrow \mathrm{Ba(NO}_{3})_{2}(aq) + 2 \mathrm{H}_{2} \mathrm{O}(\ell)\) is balanced. It shows that 2 moles of \(\mathrm{HNO}_{3}\) react with 1 mole of \(\mathrm{Ba(OH)}_{2}\). This balanced relationship is the foundation for calculating how much of each substance is required or produced.

Stoichiometric Coefficients

Stoichiometric coefficients are the numbers placed in front of compounds in chemical equations. They indicate the ratio of moles of each reactant and product involved in the reaction. These coefficients are key to solving stoichiometric problems, as they guide us in converting between moles of different substances.
For instance, in our exercise, the balanced equation \(2 \mathrm{HNO}_{3} + \mathrm{Ba(OH)}_{2} \longrightarrow \mathrm{Ba(NO}_{3})_{2} + 2 \mathrm{H}_{2} \mathrm{O} \) tells us that 2 moles of \( \mathrm{HNO}_{3} \) are needed for every 1 mole of \( \mathrm{Ba(OH)}_{2} \). By using these coefficients, we can determine how many moles of one substance are needed to fully react with or produce another substance. This is the basis for determining the required volume of solution in reactions, helping chemists and students alike predict the outcomes of their reactions accurately.