Exciting Chemistry Problem: Calculating Mass of Lithium Hydroxide

How can we calculate the mass of lithium hydroxide produced in a chemical reaction?

In the reaction of 0.900 L water at 25°C, what mass of lithium hydroxide is produced?

Step-by-Step Solution:

First, we calculate the mass of water using its density at 25°C.

Given: Density of H2O at 25°C = 0.998 g/mL

0.900 L × 0.998 g/mL = 0.8982 g

Next, we use stoichiometry to find the ratio between water and lithium hydroxide in the balanced equation:

Li + H₂O → LiOH + H₂

1 mol H2O : 1 mol LiOH

Then, we convert the mass of water to moles using the molar mass of water (18.02 g/mol):

0.8982 g H2O × (1 mol H2O / 18.02 g H2O) = 0.0499 mol H2O

After that, we convert the moles of water to moles of lithium hydroxide:

0.0499 mol H2O × (1 mol LiOH / 1 mol H2O) = 0.0499 mol LiOH

Finally, we convert the moles of lithium hydroxide to mass using the molar mass of lithium hydroxide (23.95 g/mol):

0.0499 mol LiOH × (23.95 g LiOH / 1 mol LiOH) = 1.19 g LiOH

Therefore, the mass of lithium hydroxide produced in the reaction of 0.900 L water is 1.19 grams.

Understanding the Calculation:

Chemistry problems like these involve using stoichiometry to determine the relationship between reactants and products in a chemical reaction. By balancing the equation and converting between mass, moles, and molar ratios, we can accurately calculate the mass of a specific product.

In this case, we utilized the given density of water to find the mass of water, then applied stoichiometry to establish the mole ratio between water and lithium hydroxide. Converting moles to mass allowed us to determine that 1.19 grams of lithium hydroxide would be produced in the reaction.

By following a systematic approach and understanding the principles of stoichiometry, we can solve complex chemistry problems with ease. Keep practicing and exploring different scenarios to enhance your problem-solving skills in chemistry!