Question

When barium (Ba) reacts with sulfur (S) to form barium sulfide (BaS), each Ba atom reacts with an \(\mathrm{S}\) atom. If \(2.50 \mathrm{~cm}^{3}\) of Ba reacts with \(1.75 \mathrm{~cm}^{3}\) of \(\mathrm{S},\) are there enough \(\mathrm{Ba}\) atoms to react with the \(\mathrm{S}\) atoms \(\left(d\right.\) of \(\mathrm{Ba}=3.51 \mathrm{~g} / \mathrm{cm}^{3} ; d\) of \(\left.\mathrm{S}=2.07 \mathrm{~g} / \mathrm{cm}^{3}\right) ?\)

Short answer

No, there are not enough Ba atoms.

Step-by-step solution

Calculate the mass of Barium (Ba)

Use the formula for density, which is \(\text{density} = \frac{\text{mass}}{\text{volume}}\). Rearrange to \(\text{mass} = \text{density} \times \text{volume}\). Given the density of Ba is 3.51 g/cm³ and the volume of Ba is 2.50 cm³, compute the mass of Ba: \[ \text{mass of Ba} = 3.51 \frac{\text{g}}{\text{cm}^3} \times 2.50 \text{cm}^3 = 8.775 \text{g} \]

Calculate the mass of Sulfur (S)

Similarly, use the same formula for density. Given the density of S is 2.07 g/cm³ and the volume of S is 1.75 cm³, compute the mass of S: \[ \text{mass of S} = 2.07 \frac{\text{g}}{\text{cm}^3} \times 1.75 \text{cm}^3 = 3.6225 \text{g} \]

Determine the number of moles of Barium (Ba)

Use the molar mass of Ba (137.33 g/mol) to find the number of moles. The formula is \[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \] Calculate: \[ \text{moles of Ba} = \frac{8.775 \text{g}}{137.33 \text{g/mol}} \approx 0.0639 \text{mol} \]

Determine the number of moles of Sulfur (S)

Use the molar mass of S (32.06 g/mol) in the same way to find the number of moles: \[ \text{moles of S} = \frac{3.6225 \text{g}}{32.06 \text{g/mol}} \approx 0.113 \text{mol} \]

Compare mole ratios of Ba and S

Each Ba atom reacts with one S atom to form BaS. Therefore, the mole ratio should ideally be 1:1. Compare the number of moles: \[ \text{moles of Ba} = 0.0639 \text{mol} \] and \[ \text{moles of S} = 0.113 \text{mol} \]

Conclusion

Since the moles of Ba (0.0639 mol) are less than the moles of S (0.113 mol), there are not enough Ba atoms to react with all the S atoms.

Key concepts

moles calculation

To solve stoichiometry problems, understanding moles calculation is crucial. A mole is a unit that measures the amount of a substance. It allows chemists to count atoms, molecules, or ions by weighing them.

The formula to calculate moles is: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Here,\text{mass} is the given mass of the substance, and \text{molar mass} is the mass of one mole of the substance, expressed in grams per mole.

In the given exercise, for instance, we first calculated the mass of Barium (Ba) as 8.775 g and Sulfur (S) as 3.6225 g. Using their respective molar masses, 137.33 g/mol for Ba and 32.06 g/mol for S, we then calculated their moles as follows:

\[ \text{moles of Ba} = \frac{8.775 \text{g}}{137.33 \text{g/mol}} \approx 0.0639 \text{mol} \]

\[ \text{moles of S} = \frac{3.6225 \text{g}}{32.06 \text{g/mol}} \approx 0.113 \text{mol} \]

This process requires simple division and is fundamental to determining how different substances interact in a chemical reaction.

density formula

Density is a key concept in stoichiometry and it connects the mass and volume of a substance. The formula for density is: \( \text{density} = \frac{\text{mass}}{\text{volume}} \). From here, you can rearrange this formula to find mass if you know the density and volume: \( \text{mass} = \text{density} \times \text{volume} \).

For example, in the original problem, the given densities for Barium (Ba) and Sulfur (S) were 3.51 g/cm³ and 2.07 g/cm³, respectively. By multiplying these densities by the given volumes (2.50 cm³ for Ba and 1.75 cm³ for S), we calculated their respective masses:

\[ \text{mass of Ba} = 3.51 \frac{\text{g}}{\text{cm}^3} \times 2.50 \text{cm}^3 = 8.775 \text{g} \]

\[ \text{mass of S} = 2.07 \frac{\text{g}}{\text{cm}^3} \times 1.75 \text{cm}^3 = 3.6225 \text{g} \]

This approach is straightforward and is essential for converting volume measurements into mass, which is often needed in stoichiometry problems.

molar mass

Molar mass is another fundamental concept in stoichiometry. It is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). To find the molar mass of an element, use the Atomic Mass from the periodic table. For compounds, sum the molar masses of the constituent elements.

In the given exercise, the molar mass for Barium (Ba) is 137.33 g/mol and for Sulfur (S) it is 32.06 g/mol. Using these values, we calculated the number of moles for the respective masses of Ba and S:

\[ \text{moles of Ba} = \frac{8.775 \text{g}}{137.33 \text{g/mol}} \approx 0.0639 \text{mol} \]

\[ \text{moles of S} = \frac{3.6225 \text{g}}{32.06 \text{g/mol}} \approx 0.113 \text{mol} \]

Understanding molar mass helps in converting mass to moles and vice versa, enabling us to compare quantities of substances involved in chemical reactions appropriately.