Question

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?

Short answer

2.105 moles and 319.96 grams of \(\text{Cr}_2\text{O}_3\) are required.

Step-by-step solution

Molar Mass Calculation for \(\text{Cr}_2\text{S}_3\)

Calculate the molar mass of \(\text{Cr}_2\text{S}_3\). The atomic mass of Cr (chromium) is 52.00 g/mol, and the atomic mass of S (sulfur) is 32.00 g/mol. \[ 2 \times 52.00 \text{ g/mol (Cr)} + 3 \times 32.00 \text{ g/mol (S)} = 104.00 \text{ g/mol} + 96.00 \text{ g/mol} = 200.00 \text{ g/mol} \]

Moles of \(\text{Cr}_2\text{S}_3\) Required

Determine the moles of \(\text{Cr}_2\text{S}_3\) required to produce 421 g of \(\text{Cr}_2\text{S}_3\). \[ \text{Moles of } \text{Cr}_2\text{S}_3 = \frac{421 \text{ g}}{200.00 \text{ g/mol}} = 2.105 \text{ mol } \text{Cr}_2\text{S}_3 \]

Mole Ratio from Reaction Equation

Using the balanced chemical equation, find the mole ratio between \(\text{Cr}_2\text{O}_3\) and \(\text{Cr}_2\text{S}_3\). The ratio is 1:1.

Moles of \(\text{Cr}_2\text{O}_3\) Required

Since the mole ratio is 1:1, the moles of \(\text{Cr}_2\text{O}_3\) required are the same as the moles of \(\text{Cr}_2\text{S}_3\). Therefore, 2.105 moles of \(\text{Cr}_2\text{O}_3\) are required. \[ \text{Moles of } \text{Cr}_2\text{O}_3 = 2.105 \text{ mol} \]

Molar Mass Calculation for \(\text{Cr}_2\text{O}_3\)

Calculate the molar mass of \(\text{Cr}_2\text{O}_3\). The atomic mass of Cr (chromium) is 52.00 g/mol and the atomic mass of O (oxygen) is 16.00 g/mol. \[ 2 \times 52.00 \text{ g/mol (Cr)} + 3 \times 16.00 \text{ g/mol (O)} = 104.00 \text{ g/mol} + 48.00 \text{ g/mol} = 152.00 \text{ g/mol} \]

Grams of \(\text{Cr}_2\text{O}_3\) Required

Convert the moles of \(\text{Cr}_2\text{O}_3\) required to grams using the molar mass. \[ \text{Grams of } \text{Cr}_2\text{O}_3 = 2.105 \text{ mol} \times 152.00 \text{ g/mol} = 319.96 \text{ g } \text{Cr}_2\text{O}_3 \]

Key concepts

Mole Calculation

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

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

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

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 in Chemistry

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.