(a) Outline the steps to answer the following question: What volume of [tex]\( H_2 \)[/tex] at a pressure of 745 torr and a temperature of [tex]\( 20^\circ C \)[/tex] can be prepared from the reaction of 15.0 g [tex]\( H_2O \)[/tex]?
Certainly! Let's break down the solution for the given problem step-by-step:
### (a) Outline the Steps
1. Determine the moles of H₂O: - Use the given mass of H₂O and its molar mass to find the number of moles of H₂O.
2. Determine the moles of H₂ produced: - Use the stoichiometry of the balanced chemical reaction to find the moles of H₂ that can be produced from the given moles of H₂O.
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 H₂: - 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 H₂O: - Mass of H₂O [tex]\( = 15.0 \)[/tex] grams - Molar mass of H₂O [tex]\( = 18.015 \)[/tex] g/mol - Moles of H₂O [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 H₂ produced: - According to the balanced chemical equation, 4 moles of H₂O produce 4 moles of H₂, indicating a 1:1 molar ratio. - Thus, moles of H₂ produced = moles of H₂O = 0.832 moles
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 H₂: - 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] L⋅atm/(K⋅mol) - [tex]\( T = 293.15 \)[/tex] K - Volume [tex]\( V = \frac{0.832 \text{ moles} \times 0.0821 \text{ L⋅atm/(K⋅mol)} \times 293.15 \text{ K}}{0.980 \text{ atm}} = 20.443 \text{ liters} \)[/tex]
So, the volume of [tex]\( H_2 \)[/tex] gas that can be prepared from the reaction of 15.0 g of H₂O at a pressure of 745 torr and a temperature of [tex]\( 20^{\circ} C \)[/tex] is [tex]\( 20.443 \)[/tex] liters.
