Calculate the mass of \(\mathrm{BaCO}_{3}\) produced when excess \(\mathrm{CO}_{2}\) is bubbled through a solution containing \(0.205\) moles of \(\mathrm{Ba}(\mathrm{OH})_{2}\).

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is observed. In a chemical reaction, atoms are neither created nor destroyed, so the number of atoms of each element must be the same on both sides of the equation.

Balancing begins by comparing the number of atoms of each element on both sides of the reaction. Adjustments are made by adding coefficients, which are whole numbers placed in front of compounds or elements in the equation. For our example exercise, the correct coefficients were determined to be:\begin{itemize} \begin{itemize}\begin{itemize}\begin{itemize} \begin{itemize} Balancing barium, 1 barium atom on both sides. \begin{itemize} Carbon balances with 1 carbon atom on both sides. \begin{itemize} Oxygen is balanced with 2 oxygens from hydroxide and 2 from carbon dioxide on the reactant side, and 3 from carbonate plus 1 from water on the product side. \begin{itemize} Hydrogen is balanced with 2 hydrogens from hydroxide and 2 in the water molecule. \begin{itemize} Therefore, when written out as \begin{itemize} \(\mathrm{Ba(OH)_2 + CO_2 \rightarrow BaCO_3 + H_2O}\), no coefficients are needed because the equation is already balanced with a 1:1:1:1 ratio. It is crucial to balance chemical equations accurately to proceed with stoichiometric calculations, as this determines the correct proportion of reactants and products involved.

The molar mass of a compound is the mass in grams of one mole of that substance. It is calculated by summing the masses of the individual atoms that make up the compound. Each element's atomic mass can be found on the periodic table and is measured in atomic mass units (amu).

To find the molar mass of barium carbonate \(\mathrm{BaCO_3}\), the molar masses of barium (Ba), carbon (C), and oxygen (O) are needed. The molar mass for barium is 137.33 g/mol, for carbon is 12.01 g/mol, and for oxygen is 16.00 g/mol (since there are three oxygen atoms, this is multiplied by three).

Adding these weights:\begin{itemize} \begin{itemize} 137.33 g/mol (Ba) + 12.01 g/mol (C) + 3 \(\times\) 16.00 g/mol (O) equals 197.34 g/mol for \begin{itemize} \(\mathrm{BaCO_3}\). This calculation is essential for converting between mass and moles, allowing us to use the stoichiometric ratios derived from the balanced equation to determine the mass of products or reactants in a chemical reaction. In this case, knowing the molar mass of \begin{itemize} \(\mathrm{BaCO_3}\) is crucial for the final step of our example problem.

Stoichiometry involves using the balanced chemical equation to calculate the proportions of reactants and products involved in a chemical reaction. It is based on the principle that the amounts of reactants and products in a chemical reaction are directly proportional to each other and to the coefficients of the balanced equation.

In the given example, the stoichiometric ratio between \(\mathrm{Ba(OH)_2}\) and \(\mathrm{BaCO_3}\) is 1:1. Consequently, 0.205 moles of \(\mathrm{Ba(OH)_2}\) will produce 0.205 moles of \(\mathrm{BaCO_3}\) since the reaction proceeds with a 1:1 mole ratio.

Next, this mole quantity of \(\mathrm{BaCO_3}\) is converted into mass using the calculated molar mass of 197.34 g/mol. The result is computed by the formula: \begin{itemize} \(\text{mass of \(\mathrm{BaCO_3}\)} = \text{number of moles of \(\mathrm{BaCO_3}\)} \times \text{molar mass of \(\mathrm{BaCO_3}\)}\) . The actual calculation is: \begin{itemize} \(0.205 \text{ moles} \times 197.34 \text{ g/mol} = 40.4561 \text{ g}\).

Understanding stoichiometric calculations is crucial for predicting the amounts of substances consumed and produced in a reaction, allowing for the precise design of chemical processes and ensuring that reactions proceed with the expected outcomes.