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Sagot :
Certainly! Let's break down the solution for the given problem step-by-step:
### (a) Outline the Steps
1. Determine the moles of Hâ‚‚O:
- Use the given mass of Hâ‚‚O and its molar mass to find the number of moles of Hâ‚‚O.
2. Determine the moles of Hâ‚‚ produced:
- Use the stoichiometry of the balanced chemical reaction to find the moles of Hâ‚‚ that can be produced from the given moles of Hâ‚‚O.
3. Convert pressure to appropriate units:
- Convert the given pressure from torr to atmospheres.
4. Convert temperature to appropriate units:
- Convert the given temperature from Celsius to Kelvin.
5. Use the Ideal Gas Law to find the volume of Hâ‚‚:
- Apply the Ideal Gas Law [tex]\( PV = nRT \)[/tex] to determine the volume of hydrogen gas produced.
### (b) Calculation and Answer
1. Determine the moles of Hâ‚‚O:
- Mass of Hâ‚‚O [tex]\( = 15.0 \)[/tex] grams
- Molar mass of Hâ‚‚O [tex]\( = 18.015 \)[/tex] g/mol
- Moles of Hâ‚‚O [tex]\( = \frac{\text{mass}}{\text{molar mass}} = \frac{15.0 \text{ g}}{18.015 \text{ g/mol}} = 0.832 \text{ moles} \)[/tex]
2. Determine the moles of Hâ‚‚ produced:
- According to the balanced chemical equation, 4 moles of Hâ‚‚O produce 4 moles of Hâ‚‚, indicating a 1:1 molar ratio.
- Thus, moles of Hâ‚‚ produced = moles of Hâ‚‚O = 0.832 moles
3. Convert pressure to atmospheres:
- Given pressure = 745 torr
- 1 atm = 760 torr
- Pressure in atm [tex]\( = \frac{745 \text{ torr}}{760 \text{ torr/atm}} = 0.980 \text{ atm} \)[/tex]
4. Convert temperature to Kelvin:
- Given temperature [tex]\( = 20^{\circ} C \)[/tex]
- Temperature in Kelvin [tex]\( = 20 + 273.15 = 293.15 \text{ K} \)[/tex]
5. Use the Ideal Gas Law to determine the volume of Hâ‚‚:
- Ideal Gas Law: [tex]\( PV = nRT \)[/tex]
- Rearrange to solve for [tex]\( V \)[/tex]: [tex]\( V = \frac{nRT}{P} \)[/tex]
- Where:
- [tex]\( P = 0.980 \)[/tex] atm
- [tex]\( n = 0.832 \)[/tex] moles
- [tex]\( R = 0.0821 \)[/tex] Lâ‹…atm/(Kâ‹…mol)
- [tex]\( T = 293.15 \)[/tex] K
- Volume [tex]\( V = \frac{0.832 \text{ moles} \times 0.0821 \text{ Lâ‹…atm/(Kâ‹…mol)} \times 293.15 \text{ K}}{0.980 \text{ atm}} = 20.443 \text{ liters} \)[/tex]
So, the volume of [tex]\( H_2 \)[/tex] gas that can be prepared from the reaction of 15.0 g of Hâ‚‚O at a pressure of 745 torr and a temperature of [tex]\( 20^{\circ} C \)[/tex] is [tex]\( 20.443 \)[/tex] liters.
### (a) Outline the Steps
1. Determine the moles of Hâ‚‚O:
- Use the given mass of Hâ‚‚O and its molar mass to find the number of moles of Hâ‚‚O.
2. Determine the moles of Hâ‚‚ produced:
- Use the stoichiometry of the balanced chemical reaction to find the moles of Hâ‚‚ that can be produced from the given moles of Hâ‚‚O.
3. Convert pressure to appropriate units:
- Convert the given pressure from torr to atmospheres.
4. Convert temperature to appropriate units:
- Convert the given temperature from Celsius to Kelvin.
5. Use the Ideal Gas Law to find the volume of Hâ‚‚:
- Apply the Ideal Gas Law [tex]\( PV = nRT \)[/tex] to determine the volume of hydrogen gas produced.
### (b) Calculation and Answer
1. Determine the moles of Hâ‚‚O:
- Mass of Hâ‚‚O [tex]\( = 15.0 \)[/tex] grams
- Molar mass of Hâ‚‚O [tex]\( = 18.015 \)[/tex] g/mol
- Moles of Hâ‚‚O [tex]\( = \frac{\text{mass}}{\text{molar mass}} = \frac{15.0 \text{ g}}{18.015 \text{ g/mol}} = 0.832 \text{ moles} \)[/tex]
2. Determine the moles of Hâ‚‚ produced:
- According to the balanced chemical equation, 4 moles of Hâ‚‚O produce 4 moles of Hâ‚‚, indicating a 1:1 molar ratio.
- Thus, moles of Hâ‚‚ produced = moles of Hâ‚‚O = 0.832 moles
3. Convert pressure to atmospheres:
- Given pressure = 745 torr
- 1 atm = 760 torr
- Pressure in atm [tex]\( = \frac{745 \text{ torr}}{760 \text{ torr/atm}} = 0.980 \text{ atm} \)[/tex]
4. Convert temperature to Kelvin:
- Given temperature [tex]\( = 20^{\circ} C \)[/tex]
- Temperature in Kelvin [tex]\( = 20 + 273.15 = 293.15 \text{ K} \)[/tex]
5. Use the Ideal Gas Law to determine the volume of Hâ‚‚:
- Ideal Gas Law: [tex]\( PV = nRT \)[/tex]
- Rearrange to solve for [tex]\( V \)[/tex]: [tex]\( V = \frac{nRT}{P} \)[/tex]
- Where:
- [tex]\( P = 0.980 \)[/tex] atm
- [tex]\( n = 0.832 \)[/tex] moles
- [tex]\( R = 0.0821 \)[/tex] Lâ‹…atm/(Kâ‹…mol)
- [tex]\( T = 293.15 \)[/tex] K
- Volume [tex]\( V = \frac{0.832 \text{ moles} \times 0.0821 \text{ Lâ‹…atm/(Kâ‹…mol)} \times 293.15 \text{ K}}{0.980 \text{ atm}} = 20.443 \text{ liters} \)[/tex]
So, the volume of [tex]\( H_2 \)[/tex] gas that can be prepared from the reaction of 15.0 g of Hâ‚‚O at a pressure of 745 torr and a temperature of [tex]\( 20^{\circ} C \)[/tex] is [tex]\( 20.443 \)[/tex] liters.
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