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Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is maintained. This law states that mass cannot be created or destroyed in a chemical reaction, which means the number of each type of atom must be the same on both sides of the equation.
To balance an equation, start by identifying all the reactants and products involved. For our equation:\[\mathrm{LiOH}(s) + \mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2}\mathrm{CO}_{3}(s) + \mathrm{H}_{2} \mathrm{O}(l) \]we add coefficients to balance the atoms on both sides. Place a 2 in front of \(\mathrm{LiOH}\) to achieve:\[2 \mathrm{LiOH}(s) + \mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2}\mathrm{CO}_{3}(s) + \mathrm{H}_{2}\mathrm{O}(l)\]Now, there are two lithium and two oxygen atoms on each side, making the equation balanced. Such balancing ensures we have the correct ratios to predict the amount of products formed.

Calculating molar mass is a basic yet crucial part of stoichiometry, which involves understanding the mass of one mole of a substance. It's calculated by summing the atomic masses of all the atoms in a molecular formula.
For lithium hydroxide (\(\mathrm{LiOH}\)), the molar mass calculation is:

• Lithium (Li): 6.94 g/mol
• Oxygen (O): 15.999 g/mol
• Hydrogen (H): 1.008 g/mol

Adding these gives \(23.952 \text{ g/mol for LiOH}\).
This step is critical as the moles of a substance form the basis for stoichiometric calculations. For example, converting 67.4 g of \(\mathrm{LiOH}\) into moles uses this calculated molar mass:\[\text{Moles of } \mathrm{LiOH} = \frac{67.4 \, \text{g}}{23.952 \, \text{g/mol}} = 2.815 \, \text{mol}\]These moles are then utilized in further stoichiometric calculations.

Stoichiometric coefficients are the numbers placed in front of compounds in a balanced chemical equation. They indicate the relative number of moles of each reactant and product involved in the reaction. This allows us to predict how much of one substance will react or be produced by the reaction.
In our balanced equation:\[2 \mathrm{LiOH}(s) + \mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2}\mathrm{CO}_{3}(s) + \mathrm{H}_{2}\mathrm{O}(l)\]The stoichiometric coefficients are 2, 1, 1, and 1 for \(\mathrm{LiOH}\), \(\mathrm{CO}_{2}\), \(\mathrm{Li}_{2}\mathrm{CO}_{3}\), and water respectively.
These coefficients guide the mole ratios used in stoichiometry. For instance, 2 moles of \(\mathrm{LiOH}\) produce 1 mole of \(\mathrm{Li}_{2}\mathrm{CO}_{3}\). Thus, if we start with 2.815 moles of \(\mathrm{LiOH}\), we calculate moles of \(\mathrm{Li}_{2}\mathrm{CO}_{3}\) produced using the ratio:\[\text{Moles of } \mathrm{Li}_{2}\mathrm{CO}_{3} = \frac{1}{2} \times 2.815 = 1.4075 \, \text{mol}\]Understanding these coefficients is essential for the accuracy in predicting the yields of reactions.