Question

Consider the following unbalanced chemical equation. $$ \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ If \(67.4 \mathrm{~g}\) of lithium hydroxide reacts with excess carbon dioxide, what mass of lithium carbonate will be produced?

Short answer

By balancing the chemical equation and performing stoichiometric calculations, we find that approximately 103.97 g of lithium carbonate (Li₂CO₃) will be produced when 67.4 g of lithium hydroxide (LiOH) reacts with excess carbon dioxide (CO₂).

Step-by-step solution

Balance the chemical equation

Balance the given chemical equation using stoichiometric coefficients: $$ 2 \mathrm{LiOH}(s)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{Li}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$

Convert the mass of LiOH to moles

For the balanced equation, we need to convert the mass of LiOH given (67.4 g) into moles. To do this, we use the molecular weight (molar mass) of LiOH: 1(6.94 g/mol) + 1(15.999 g/mol) + 1(1.008 g/mol) = 23.952 g/mol LiOH. $$ \text{Moles of LiOH} = \frac{\text{mass of LiOH}}{\text{molar mass of LiOH}} = \frac{67.4 \mathrm{~g}}{23.952 \mathrm{~g/mol}} = 2.815 \mathrm{~mol} \text{ LiOH} $$

Use stoichiometry to find moles of Li₂CO₃ produced

According to the stoichiometry of the balanced equation, 2 moles of LiOH react to produce 1 mole of Li₂CO₃. $$ \text{Moles of Li₂CO₃} = \frac{1 \mathrm{~mol} \text{ Li₂CO₃}}{2 \mathrm{~mol} \text{ LiOH}} \times 2.815 \mathrm{~mol} \text{ LiOH} = 1.4075 \mathrm{~mol} \text{ Li₂CO₃} $$

Convert moles of Li₂CO₃ to mass

Now we need to convert the moles of Li₂CO₃ produced to mass, using its molecular weight (molar mass): 2(6.94 g/mol) + 1(12.011 g/mol) + 3(15.999 g/mol) = 73.891 g/mol Li₂CO₃. $$ \text{Mass of Li₂CO₃} = \text{moles of Li₂CO₃} \times \text{molar mass of Li₂CO₃} = 1.4075 \mathrm{~mol} \text{ Li₂CO₃} \times 73.891 \mathrm{~g/mol} = 103.97 \mathrm{~g} \text{ Li₂CO₃} $$ So, the mass of lithium carbonate (Li₂CO₃) that will be produced is approximately 103.97 g.

Key concepts

Balancing Chemical Equations

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.

Molar Mass Calculation

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

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.