Question

What mass of \(\mathrm{BaSO}_{4}\) will be precipitated from a large container of concentrated \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) solution if \(37.5 \mathrm{mL}\) of \(0.221 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is added?

Short answer

The mass of precipitated \(\mathrm{BaSO}_{4}\) can be calculated using the given volume and molarity of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) and stoichiometry of the reaction. Firstly, we find the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) by multiplying its volume (0.0375 L) and molarity (0.221 M). Next, based on the balanced chemical reaction, the moles of \(\mathrm{BaSO}_{4}\) is equal to the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\). Finally, to calculate the mass of \(\mathrm{BaSO}_{4}\), multiply the moles of \(\mathrm{BaSO}_{4}\) by its molar mass (233.40 g/mol).

Step-by-step solution

Write the balanced chemical reaction

First, we need to write the balanced chemical reaction between \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{H}_{2}\mathrm{SO}_{4}\). The chemical reaction is: \[ \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} + \mathrm{H}_{2}\mathrm{SO}_4 \rightarrow \mathrm{BaSO}_{4} + 2 \mathrm{HNO}_{3} \]

Calculate moles of sulfuric acid

Next, we need to calculate the moles of sulfuric acid, \(\mathrm{H}_{2}\mathrm{SO}_{4}\), using the given volume and molarity. The formula to find moles is: Moles = Volume x Molarity We know: - Volume of sulfuric acid = 37.5 mL = 0.0375 L (convert mL to L) - Molarity of sulfuric acid = 0.221 M So, the moles of sulfuric acid can be calculated as follows: Moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) = 0.0375 L x 0.221 M

Calculate moles of precipitate

Based on the stoichiometry of the balanced chemical reaction, we can see that: 1 mole of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) reacts with 1 mole of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) to produce 1 mole of \(\mathrm{BaSO}_{4}\). Since we have already calculated the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\), we can now find the moles of \(\mathrm{BaSO}_{4}\). In this case, it is equal to the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\): Moles of \(\mathrm{BaSO}_{4}\) = Moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\)

Calculate mass of precipitate

Finally, we can calculate the mass of \(\mathrm{BaSO}_{4}\) precipitated using its molar mass: Mass = Moles x Molar mass Molar mass of \(\mathrm{BaSO}_{4}\) = \(\mathrm{Ba}\) (137.33 g/mol) + \(\mathrm{S}\) (32.07 g/mol) + \(\mathrm{O}_{4}\) (4 x 16.00 g/mol) = 233.40 g/mol We can now find the mass of \(\mathrm{BaSO}_{4}\): Mass of \(\mathrm{BaSO}_{4}\) = Moles of \(\mathrm{BaSO}_{4}\) x Molar mass of \(\mathrm{BaSO}_{4}\) Plug in the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) from Step 2, and then calculate the mass of \(\mathrm{BaSO}_{4}\).

Key concepts

Chemical Reaction Balancing

Understanding how to balance chemical reactions is crucial in chemistry, as it reflects the conservation of mass and the stoichiometry of the reaction. To balance a chemical equation, we must make sure that the number of atoms of each element on the reactant side equals the number on the product side. In our precipitation reaction, the balanced equation is:
\[ \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} + \mathrm{H}_{2}\mathrm{SO}_4 \rightarrow \mathrm{BaSO}_{4} + 2 \mathrm{HNO}_{3} \]
Balancing involves the addition of coefficients — in this case, 1 for \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{H}_{2}\mathrm{SO}_4\), and 2 for \(\mathrm{HNO}_{3}\). Every atom accounted for in the reactants now appears in the same number in the products.

Mole Concept

The mole concept is a bridge between the micro world of atoms and molecules and the macro world of grams and liters. One mole is defined as \(6.022 \times 10^{23}\) entities (Avogadro's number) of anything - be that atoms, molecules, ions, or electrons. It's a counting unit much like a 'dozen', but for a very large number of particles.
In our exercise, we calculate the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) by using the formula:
\[ \text{Moles} = \text{Volume} \times \text{Molarity} \]
This calculation gives us the number of moles of sulfuric acid, which is then used to find the amount of \(\mathrm{BaSO}_{4}\) that will precipitate.

Molarity Calculations

Molarity, denoted as \(M\), is a measure of concentration in chemistry, indicating the number of moles of a solute present in one liter of solution. The formula for calculating molarity is:
\[ \text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in liters}} \]
In our scenario, to find the moles of \(\mathrm{H}_{2}\mathrm{SO}_{4}\) used, we multiply the volume of the acid (in liters) by its molarity. Once we have the moles of acid, we can use stoichiometry to determine the moles of Barium Sulfate that will form.

Stoichiometry

Stoichiometry comes from the Greek words for 'element' and 'measure', and it deals with the quantification of reactants and products in a chemical reaction. It is based on the balanced equation and uses the mole concept to convert between masses, volumes, and numbers of particles.
In a balanced reaction, the stoichiometry tells us the ratio in which the reactants combine and the products form. For the precipitation of \(\mathrm{BaSO}_{4}\), we see that it is a 1:1 stoichiometric ratio with \(\mathrm{H}_{2}\mathrm{SO}_{4}\). Thus, we can predict the moles of precipitate formed from the calculated moles of sulfuric acid. To find the mass of the precipitate, we multiply its moles by its molar mass, thus converting the abstract mole concept into a tangible mass that can be measured in the lab.