Chromium(III) oxide reacts with hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) gas to form chromium(III) sulfide and water: $$ \mathrm{Cr}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2} \mathrm{~S}(g) \longrightarrow \mathrm{Cr}_{2} \mathrm{~S}_{3}(s)+3 \mathrm{H}_{2} \mathrm{O}(l) $$ To produce \(421 \mathrm{~g}\) of \(\mathrm{Cr}_{2} \mathrm{~S}_{3},\) how many (a) moles of \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) and (b) grams of \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) are required?

Mole calculation is a fundamental concept in chemistry. It allows you to quantify the amount of a substance. One mole is equivalent to Avogadro's number \(6.022 \times 10^{23}\) of particles (atoms, molecules, ions). For example, in the given exercise, to produce a specific amount of Chromium(III) sulfide (\text{Cr}_2\text{S}_3\text{), we calculated the moles needed based on its given mass (421 g) and molar mass (200 g/mol): \(2.105 \text{ mol } \text{Cr}_2\text{S}_3\). This ensures that we have the right quantity needed for the reaction. Always ensure to convert the mass into moles for precise stoichiometric calculations.

Stoichiometry involves dealing with the quantitative relationships between reactants and products in a chemical reaction. For the reaction between Chromium(III) oxide (\text{Cr}_2\text{O}_3\text{) and Hydrogen sulfide (\text{H}_2\text{S}\text{), we use stoichiometric ratios derived from the balanced equation: \( \text{Cr}_{2} \text{O}_3(s) + 3 \text{H}_2 \text{S}(g) \rightarrow \text{Cr}_{2} \text{S}_{3}(s) + 3 \text{H}_2 \text{O}(l) \). From this equation, a 1:1 mole ratio between \text{Cr}_2\text{O}_3\text{ and \text{Cr}_2\text{S}_3\text{ can be derived. This means, for every mole of \text{Cr}_2\text{O}_3\text{ used, one mole of \text{Cr}_2\text{S}_3\text{ is produced, which was crucial in determining that 2.105 mol of \text{Cr}_2\text{O}_3\text{ is needed to produce 421 g of \text{Cr}_2\text{S}_3\text).

Chemical reactions involve the transformation of reactants into products. Understanding the dynamics of these reactions helps in predicting the amounts of substances consumed and formed. In this exercise, the reaction between Chromium(III) oxide and Hydrogen sulfide results in the formation of Chromium(III) sulfide and water. The balanced chemical equation provides insights into the precise relationships and quantities involved. Once balanced, it serves as the roadmap for all mole and mass calculations needed in stoichiometry.

Molar mass calculation is essential to convert between mass and moles. The molar mass is the mass of one mole of a substance, expressed in g/mol. For instance, the molar mass of Chromium(III) sulfide (\text{Cr}_2\text{S}_3\text{) was calculated as: \(2 \times 52.00 \text{ g/mol (Cr)} + 3 \times 32.00 \text{ g/mol (S)} = 200.00 \text{ g/mol}\). Similarly, for \text{Cr}_2\text{O}_3\text{, the molar mass is: \(2 \times 52.00 \text{ g/mol (Cr)} + 3 \times 16.00 \text{ g/mol (O)} = 152.00 \text{ g/mol}\). With these values, we can convert between grams and moles, aiding the solution of the exercise.

Conversion of units is a critical skill in chemistry to move seamlessly between different measurements. In this exercise, we converted mass to moles and vice versa using molar masses. First, we found the moles of \text{Cr}_2\text{S}_3\text{ required from its mass: \( \text{Moles of } \text{Cr}_2\text{S}_3 = \frac{421 \text{ g}}{200.00 \text{ g/mol}} = 2.105 \text{ mol } \text{Cr}_2\text{S}_3 \). Then, using stoichiometry, we determined that 2.105 moles of \text{Cr}_2\text{O}_3\text{ were needed. Finally, we converted moles of \text{Cr}_2\text{O}_3\text{ back to grams, yielding \( \text{Grams of } \text{Cr}_2\text{O}_3 = 2.105 \text{ mol} \times 152.00 \text{ g/mol} = 319.96 \text{ g } \text{Cr}_2\text{O}_3 \). Each of these steps involves critical unit conversion, emphasizing the importance of this skill in solving chemical problems accurately.