Let's solve this step by step:
### Step 1: Convert Volume to Liters
The volume of hydrogen gas collected is given as 511 mL. To convert this to liters, we use the conversion factor [tex]\(1 \text{ L} = 1000 \text{ mL}\)[/tex].
[tex]\[ \text{Volume of }\text{H}_2 = 511 \text{ mL} \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.511 \text{ L} \][/tex]
### Step 2: Calculate Pressure of Hydrogen Gas
The total pressure measured is 745 mmHg. Since the gas is collected over water, we need to account for the vapor pressure of water, which is 24 mmHg at 26°C.
[tex]\[ \text{Pressure of }\text{H}_2 = 745 \text{ mmHg} - 24 \text{ mmHg} = 721 \text{ mmHg} \][/tex]
### Step 3: Convert Pressure to Atmospheres
The pressure in mmHg needs to be converted to atmospheres using the conversion factor [tex]\(1 \text{ atm} = 760 \text{ mmHg}\)[/tex].
[tex]\[ \text{Pressure of }\text{H}_2 = \frac{721 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.9487 \text{ atm} \][/tex]
### Step 4: Convert Temperature to Kelvin
Temperature needs to be converted from Celsius to Kelvin using the conversion [tex]\( K = \text{°C} + 273.15 \)[/tex].
[tex]\[ \text{Temperature in Kelvin} = 26 \text{°C} + 273.15 = 299.15 \text{ K} \][/tex]
### Step 5: Calculate Moles of Hydrogen Gas ([tex]\(n_{\text{H}_2}\)[/tex])
Using the ideal gas law [tex]\( PV = nRT \)[/tex], where:
- [tex]\( P \)[/tex] is the pressure in atmospheres (0.9487 atm),
- [tex]\( V \)[/tex] is the volume in liters (0.511 L),
- [tex]\( R \)[/tex] is the gas constant (0.0821 L·atm/(mol·K)),
- [tex]\( T \)[/tex] is the temperature in Kelvin (299.15 K).
Rearranging the ideal gas law to solve for [tex]\( n \)[/tex]:
[tex]\[ n_{\text{H}_2} = \frac{PV}{RT} = \frac{0.9487 \text{ atm} \times 0.511 \text{ L}}{0.0821 \text{ L·atm/(mol·K)} \times 299.15 \text{ K}} = 0.01974 \text{ moles}\][/tex]
### Step 6: Determine Moles of Aluminum Reacted
From the balanced chemical equation:
[tex]\[ 2 \text{Al} + 6 \text{HCl} \rightarrow 2 \text{AlCl}_3 + 3 \text{H}_2 \][/tex]
The molar ratio of [tex]\(\text{Al}\)[/tex] to [tex]\(\text{H}_2\)[/tex] is 2:3. Therefore, for every 3 moles of [tex]\(\text{H}_2\)[/tex], 2 moles of [tex]\(\text{Al}\)[/tex] are reacted:
[tex]\[ n_{\text{Al}} = \frac{2}{3} \times n_{\text{H}_2} = \frac{2}{3} \times 0.01974 \text{ moles} = 0.01316 \text{ moles}\][/tex]
### Step 7: Calculate Mass of Aluminum Reacted
The molar mass of aluminum (Al) is 26.98 g/mol. To find the mass of aluminum reacted:
[tex]\[ \text{Mass of }\text{Al} = n_{\text{Al}} \times \text{Molar mass of }\text{Al} = 0.01316 \text{ moles} \times 26.98 \text{ g/mol} = 0.3550 \text{ grams}\][/tex]
So, the mass of aluminum that reacted is approximately 0.355 grams.
