(1 point) How many moles are contained in 23.5 g of Sb2S3? 2. (1 point) How many formula units are contained in 23.5 g of Sb2S3? 3. (1 point) How many sulfide ions are present in 23.5 g of Sb2S3? 4.(1 point) Given 5.0 x 1024 K2SO4 formula units, how many moles are present? 5.(1 point) Given 5.0 x 1024 K2SO4 formula units, how many grams are present?
The Correct Answer and Explanation is :
1. How many moles are contained in 23.5 g of Sb₂S₃?
To calculate the moles of Sb₂S₃, we first need to determine the molar mass of Sb₂S₃. The molar mass can be calculated using the atomic masses of antimony (Sb) and sulfur (S).
- Atomic mass of Sb = 121.76 g/mol
- Atomic mass of S = 32.06 g/mol
Molar mass of Sb₂S₃ = (2 × 121.76 g/mol) + (3 × 32.06 g/mol)
= 243.52 g/mol + 96.18 g/mol
= 339.70 g/mol
Now, we can calculate the number of moles in 23.5 g of Sb₂S₃:
[
\text{moles of Sb₂S₃} = \frac{\text{mass of Sb₂S₃}}{\text{molar mass of Sb₂S₃}} = \frac{23.5 \, \text{g}}{339.70 \, \text{g/mol}} = 0.0692 \, \text{mol}
]
Thus, there are 0.0692 moles of Sb₂S₃ in 23.5 g.
2. How many formula units are contained in 23.5 g of Sb₂S₃?
The number of formula units is calculated using Avogadro’s number ((6.022 \times 10^{23}) units/mol). To find the number of formula units, multiply the number of moles by Avogadro’s number:
[
\text{formula units} = 0.0692 \, \text{mol} \times 6.022 \times 10^{23} \, \text{units/mol} = 4.17 \times 10^{22} \, \text{formula units}
]
Thus, there are 4.17 × 10²² formula units of Sb₂S₃ in 23.5 g.
3. How many sulfide ions are present in 23.5 g of Sb₂S₃?
Each formula unit of Sb₂S₃ contains 3 sulfide ions (S²⁻). To find the number of sulfide ions, we multiply the number of formula units by 3:
[
\text{sulfide ions} = 3 \times 4.17 \times 10^{22} = 1.25 \times 10^{23} \, \text{ions}
]
Thus, there are 1.25 × 10²³ sulfide ions in 23.5 g of Sb₂S₃.
4. Given 5.0 × 10²⁴ K₂SO₄ formula units, how many moles are present?
To calculate moles, use the formula:
[
\text{moles of K₂SO₄} = \frac{\text{formula units of K₂SO₄}}{\text{Avogadro’s number}} = \frac{5.0 \times 10^{24}}{6.022 \times 10^{23}} = 8.3 \, \text{mol}
]
Thus, there are 8.3 moles of K₂SO₄.
5. Given 5.0 × 10²⁴ K₂SO₄ formula units, how many grams are present?
First, calculate the molar mass of K₂SO₄:
- Atomic mass of K = 39.10 g/mol
- Atomic mass of S = 32.06 g/mol
- Atomic mass of O = 16.00 g/mol
Molar mass of K₂SO₄ = (2 × 39.10) + 32.06 + (4 × 16.00) = 78.20 + 32.06 + 64.00 = 174.26 g/mol
Now, convert moles to grams:
[
\text{mass of K₂SO₄} = 8.3 \, \text{mol} \times 174.26 \, \text{g/mol} = 1443.35 \, \text{g}
]
Thus, there are 1443.35 grams of K₂SO₄ in 5.0 × 10²⁴ formula units.
Summary of answers:
- 0.0692 moles of Sb₂S₃
- 4.17 × 10²² formula units of Sb₂S₃
- 1.25 × 10²³ sulfide ions in Sb₂S₃
- 8.3 moles of K₂SO₄
- 1443.35 grams of K₂SO₄