Answer :
Sure, let's go through the given problem step by step:
(a) Calculate the partial pressure of oxygen in the gas collected:
1. Determine the total pressure and the vapor pressure of water:
- Total pressure of the gas collected [tex]\( P_{\text{total}} \)[/tex] = 754 torr
- Vapor pressure of water at [tex]\( 22^{\circ} \text{C} \)[/tex] [tex]\( P_{\text{H}_2\text{O}} \)[/tex] = 21 torr
2. Apply Dalton's Law of Partial Pressures to find the partial pressure of oxygen ([tex]\( P_{\text{O}_2} \)[/tex]):
[tex]\[ P_{\text{O}_2} = P_{\text{total}} - P_{\text{H}_2\text{O}} \][/tex]
Substituting the known values:
[tex]\[ P_{\text{O}_2} = 754 \, \text{torr} - 21 \, \text{torr} = 733 \, \text{torr} \][/tex]
So, the partial pressure of oxygen in the gas collected is 733 torr.
(b) Calculate how many grams of [tex]\( \text{KClO}_3 \)[/tex] were decomposed:
1. Convert the partial pressure of oxygen from torr to atm:
[tex]\[ P_{\text{O}_2} (\text{atm}) = \frac{733 \, \text{torr}}{760 \, \text{torr/atm}} = 0.9645 \, \text{atm} \][/tex]
2. Use the ideal gas law to find the moles of [tex]\( \text{O}_2 \)[/tex]:
- Given the volume [tex]\( V = 0.65 \, \text{L} \)[/tex]
- Temperature [tex]\( T = 22^\circ \text{C} \)[/tex] which needs to be converted to Kelvin:
[tex]\[ T (\text{K}) = 22 + 273.15 = 295.15 \, \text{K} \][/tex]
- Ideal gas constant [tex]\( R = 0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}) \)[/tex]
The ideal gas law is [tex]\( PV = nRT \)[/tex], solving for [tex]\( n \)[/tex]:
[tex]\[ n_{\text{O}_2} = \frac{P_{\text{O}_2} \cdot V}{R \cdot T} = \frac{0.9645 \, \text{atm} \cdot 0.65 \, \text{L}}{0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}) \cdot 295.15 \, \text{K}} = 0.02587 \, \text{mol} \][/tex]
3. Determine the moles of [tex]\( \text{KClO}_3 \)[/tex] decomposed using stoichiometry:
From the balanced equation:
[tex]\[ 2 \, \text{KClO}_3 (s) \rightarrow 2 \, \text{KCl} (s) + 3 \, \text{O}_2 (g) \][/tex]
The mole ratio of [tex]\( \text{KClO}_3 \)[/tex] to [tex]\( \text{O}_2 \)[/tex] is 2:3. Therefore:
[tex]\[ \frac{2 \, \text{mol} \, \text{KClO}_3}{3 \, \text{mol} \, \text{O}_2} \times 0.02587 \, \text{mol} \, \text{O}_2 = 0.01725 \, \text{mol} \, \text{KClO}_3 \][/tex]
4. Finally, calculate the mass of [tex]\( \text{KClO}_3 \)[/tex] decomposed:
- The molar mass of [tex]\( \text{KClO}_3 \)[/tex] is calculated as follows:
[tex]\[ \text{K}: 39.1 \, \text{g/mol}, \quad \text{Cl}: 35.5 \, \text{g/mol}, \quad \text{O}: 16 \, \text{g/mol} \times 3 = 48 \, \text{g/mol} \][/tex]
So,
[tex]\[ \text{Molar mass of } \text{KClO}_3 = 39.1 + 35.5 + 48 = 122.6 \, \text{g/mol} \][/tex]
Using the moles of [tex]\( \text{KClO}_3 \)[/tex] decomposed:
[tex]\[ \text{Mass of } \text{KClO}_3 = 0.01725 \, \text{mol} \times 122.6 \, \text{g/mol} = 2.1137 \, \text{g} \][/tex]
So, the mass of [tex]\( \text{KClO}_3 \)[/tex] decomposed is 2.114 grams.
In summary:
- The partial pressure of oxygen in the gas collected is 733 torr.
- The mass of [tex]\( \text{KClO}_3 \)[/tex] decomposed is 2.114 grams.
(a) Calculate the partial pressure of oxygen in the gas collected:
1. Determine the total pressure and the vapor pressure of water:
- Total pressure of the gas collected [tex]\( P_{\text{total}} \)[/tex] = 754 torr
- Vapor pressure of water at [tex]\( 22^{\circ} \text{C} \)[/tex] [tex]\( P_{\text{H}_2\text{O}} \)[/tex] = 21 torr
2. Apply Dalton's Law of Partial Pressures to find the partial pressure of oxygen ([tex]\( P_{\text{O}_2} \)[/tex]):
[tex]\[ P_{\text{O}_2} = P_{\text{total}} - P_{\text{H}_2\text{O}} \][/tex]
Substituting the known values:
[tex]\[ P_{\text{O}_2} = 754 \, \text{torr} - 21 \, \text{torr} = 733 \, \text{torr} \][/tex]
So, the partial pressure of oxygen in the gas collected is 733 torr.
(b) Calculate how many grams of [tex]\( \text{KClO}_3 \)[/tex] were decomposed:
1. Convert the partial pressure of oxygen from torr to atm:
[tex]\[ P_{\text{O}_2} (\text{atm}) = \frac{733 \, \text{torr}}{760 \, \text{torr/atm}} = 0.9645 \, \text{atm} \][/tex]
2. Use the ideal gas law to find the moles of [tex]\( \text{O}_2 \)[/tex]:
- Given the volume [tex]\( V = 0.65 \, \text{L} \)[/tex]
- Temperature [tex]\( T = 22^\circ \text{C} \)[/tex] which needs to be converted to Kelvin:
[tex]\[ T (\text{K}) = 22 + 273.15 = 295.15 \, \text{K} \][/tex]
- Ideal gas constant [tex]\( R = 0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}) \)[/tex]
The ideal gas law is [tex]\( PV = nRT \)[/tex], solving for [tex]\( n \)[/tex]:
[tex]\[ n_{\text{O}_2} = \frac{P_{\text{O}_2} \cdot V}{R \cdot T} = \frac{0.9645 \, \text{atm} \cdot 0.65 \, \text{L}}{0.0821 \, \text{L} \cdot \text{atm} / (\text{mol} \cdot \text{K}) \cdot 295.15 \, \text{K}} = 0.02587 \, \text{mol} \][/tex]
3. Determine the moles of [tex]\( \text{KClO}_3 \)[/tex] decomposed using stoichiometry:
From the balanced equation:
[tex]\[ 2 \, \text{KClO}_3 (s) \rightarrow 2 \, \text{KCl} (s) + 3 \, \text{O}_2 (g) \][/tex]
The mole ratio of [tex]\( \text{KClO}_3 \)[/tex] to [tex]\( \text{O}_2 \)[/tex] is 2:3. Therefore:
[tex]\[ \frac{2 \, \text{mol} \, \text{KClO}_3}{3 \, \text{mol} \, \text{O}_2} \times 0.02587 \, \text{mol} \, \text{O}_2 = 0.01725 \, \text{mol} \, \text{KClO}_3 \][/tex]
4. Finally, calculate the mass of [tex]\( \text{KClO}_3 \)[/tex] decomposed:
- The molar mass of [tex]\( \text{KClO}_3 \)[/tex] is calculated as follows:
[tex]\[ \text{K}: 39.1 \, \text{g/mol}, \quad \text{Cl}: 35.5 \, \text{g/mol}, \quad \text{O}: 16 \, \text{g/mol} \times 3 = 48 \, \text{g/mol} \][/tex]
So,
[tex]\[ \text{Molar mass of } \text{KClO}_3 = 39.1 + 35.5 + 48 = 122.6 \, \text{g/mol} \][/tex]
Using the moles of [tex]\( \text{KClO}_3 \)[/tex] decomposed:
[tex]\[ \text{Mass of } \text{KClO}_3 = 0.01725 \, \text{mol} \times 122.6 \, \text{g/mol} = 2.1137 \, \text{g} \][/tex]
So, the mass of [tex]\( \text{KClO}_3 \)[/tex] decomposed is 2.114 grams.
In summary:
- The partial pressure of oxygen in the gas collected is 733 torr.
- The mass of [tex]\( \text{KClO}_3 \)[/tex] decomposed is 2.114 grams.
