Question

The mineral stibnite, antimony(III) sulfide, is treated with hydrochloric acid to give antimony(III) chloride and hydrogen sulfide gas. What STP volume of \(\mathrm{H}_{2} \mathrm{~S}\) is produced from a 3.00 -g sample of stibnite?

Short answer

0.593 L of \( \text{H}_2\text{S} \) is produced from a 3.00 g sample of stibnite.

Step-by-step solution

Write the balanced chemical equation

First, we need to determine the balanced chemical equation for this reaction. The mineral stibnite, with the formula \( \text{Sb}_2\text{S}_3 \), reacts with \( \text{HCl} \) to produce \( \text{SbCl}_3 \) and \( \text{H}_2\text{S} \). The balanced chemical equation is:\[\text{Sb}_2\text{S}_3 (s) + 6 \text{HCl} (aq) \rightarrow 2 \text{SbCl}_3 (aq) + 3 \text{H}_2\text{S} (g)\]

Calculate moles of stibnite (Sb extsubscript{2}S extsubscript{3})

Find the molar mass of stibnite, \( \text{Sb}_2\text{S}_3 \):- Antimony (Sb): 121.76 g/mol- Sulfur (S): 32.07 g/molMolar mass of \( \text{Sb}_2\text{S}_3 = 2(121.76) + 3(32.07) = 339.62 \text{ g/mol} \).Using the given mass (3.00 g), calculate moles:\[\text{Moles of } \text{Sb}_2\text{S}_3 = \frac{3.00 \text{ g}}{339.62 \text{ g/mol}} \approx 0.00883 \text{ mol}\]

Determine moles of H extsubscript{2}S produced

According to the balanced equation, 1 mole of \( \text{Sb}_2\text{S}_3 \) produces 3 moles of \( \text{H}_2\text{S} \).Thus, the moles of \( \text{H}_2\text{S} \) produced are:\[0.00883 \text{ mol of } \text{Sb}_2\text{S}_3 \times \frac{3 \text{ mol } \text{H}_2\text{S}}{1 \text{ mol } \text{Sb}_2\text{S}_3} = 0.02649 \text{ mol } \text{H}_2\text{S}\]

Convert moles of H extsubscript{2}S to volume at STP

At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Therefore, the volume of \( \text{H}_2\text{S} \) produced is:\[0.02649 \text{ mol } \text{H}_2\text{S} \times 22.4 \text{ L/mol} = 0.5934 \text{ L}\]

Solution Conclusion

Based on the calculations above, the STP volume of \( \text{H}_2\text{S} \) gas produced from a 3.00 g sample of stibnite is approximately 0.593 L.

Key concepts

Understanding Chemical Equations

Chemical equations are fundamental in stoichiometry, serving as the blueprint for a chemical reaction. Each component in a chemical equation represents a substance, either an element or a compound, participating in the reaction. In the case of our original exercise, the balanced equation is:

• \[\text{Sb}_2\text{S}_3 (s) + 6 \text{HCl} (aq) \rightarrow 2 \text{SbCl}_3 (aq) + 3 \text{H}_2\text{S} (g)\]

This equation tells us that one mole of stibnite (\(\text{Sb}_2\text{S}_3\)) reacts with six moles of hydrochloric acid (\(\text{HCl}\)) to produce two moles of antimony chloride (\(\text{SbCl}_3\)) and three moles of hydrogen sulfide (\(\text{H}_2\text{S}\)). A balanced equation ensures conservation of mass, meaning the number of atoms of each element is the same on both sides of the equation, satisfying the law of conservation of matter.
To balance the equation, you adjust the coefficients (the numbers in front of the chemicals) to ensure that the elements are balanced. This involves trial and error and a solid understanding of chemical formulas.
Balancing chemical equations is an essential skill in stoichiometry, allowing you to understand the quantitative relationships between reactants and products.

Calculating Molar Mass

The molar mass of a substance is a crucial factor when it comes to calculating the amount of a substance (in moles) that is involved in a chemical reaction. It's essentially the mass of one mole of a given substance, expressed in grams per mole (g/mol).
In our given reaction, the molar mass of stibnite (\(\text{Sb}_2\text{S}_3\)) is calculated by adding together the mass of all the atoms in the molecule:

• Antimony, \(\text{Sb}\), has a molar mass of 121.76 g/mol.
• Sulfur, \(\text{S}\), has a molar mass of 32.07 g/mol.
• Therefore, the molar mass of \(\text{Sb}_2\text{S}_3\) is:\[2(121.76 \, \text{g/mol}) + 3(32.07 \, \text{g/mol}) = 339.62 \, \text{g/mol}\]

Accurately calculating the molar mass allows us to convert between grams and moles, which is essential when predicting how much product a reaction will produce based on a given amount of reactant. This forms the core of stoichiometric calculations and aids in making precise measurements in a laboratory setting.

Gas Laws and Volume Calculations at STP

When dealing with gases in chemical reactions, understanding the principles outlined by gas laws is essential, particularly when calculating the volume of gas produced or consumed. At standard temperature and pressure (STP)—defined as 0°C (273.15 K) and 1 atm pressure—1 mole of any ideal gas occupies a volume of 22.4 liters.
In the exercise, after determining the moles of hydrogen sulfide (\(\text{H}_2\text{S}\)) produced, the next step is to find its volume at STP. The calculation uses the molar volume of a gas at STP, which is a consistent value for all ideal gases:

• The moles of \(\text{H}_2\text{S}\) were calculated to be 0.02649 mol.
• Using the molar volume at STP:
  • \[0.02649 \, \text{mol} \times 22.4 \, \text{L/mol} = 0.5934 \, \text{L}\]

Thus, at STP, the \(\text{H}_2\text{S}\) gas produced occupies a volume of approximately 0.593 liters. Utilizing the concept of molar volume at STP simplifies the process of calculating the gas volumes involved in reactions, which is an essential step in analyzing and designing chemical processes.