Question

How many grams of cl2 are in a 0.500 liter sample at stop?

Answer 1

To calculate the mass of Cl2 in a 0.500 L sample at STP, calculate the moles of Cl2 and multiply by its molar mass. The sample contains approximately 1.583 grams of Cl2.

Step-by-step explanation:

The student's question pertains to determining the mass of chlorine gas (Cl2) contained in a sample. To calculate this, we must first use the conversion factor that relates moles of chlorine to its mass. This is given as 1 mole of Cl2 is equivalent to 70.90 grams of Cl2. Assuming the sample is at standard temperature and pressure (STP), one mole of any gas occupies 22.4 liters. Therefore, we can calculate the number of moles of Cl2 in 0.500 liters, and then use that number of moles to find the mass in grams.

Step 1: Calculate the moles of Cl2 present in 0.500 liters at STP:

• (0.500 L Cl2) / (22.4 L/mol) = 0.02232 mol Cl2

Step 2: Convert moles of Cl2 to grams:

• (0.02232 mol Cl2) * (70.90 g/mol) = 1.583 g Cl2

Therefore, the 0.500-liter sample of Cl2 at STP contains approximately 1.583 grams of Cl2.

Answer 2

The number of grams of Cl2 in a 0.500-liter sample at STP is approximately 1.121 grams.

Step-by-step explanation:

Chlorine gas (Cl2) at Standard Temperature and Pressure (STP) conditions has a molar volume of 22.4 liters/mol. This means that one mole of Cl2 occupies 22.4 liters at STP. To find the number of moles of Cl2 in a 0.500-liter sample, we can use the relationship:

`\[ \text{moles} = \frac{\text{volume (liters)}}{\text{molar volume (liters/mol)}} \]`

Substituting the given values:

`\[ \text{moles of Cl2} = \frac{0.500 \, \text{liters}}{22.4 \, \text{liters/mol}} \]`

`\[ \text{moles of Cl2} \approx 0.0223 \, \text{mol} \]`

Next, we use the molar mass of chlorine (Cl), which is approximately 35.453 g/mol, to find the mass of Cl2:

`\[ \text{mass} = \text{moles} * \text{molar mass} \]`

`\[ \text{mass of Cl2} \approx 0.0223 \, \text{mol} * 35.453 \, \text{g/mol} \]`

`\[ \text{mass of Cl2} \approx 1.121 \, \text{grams} \]`

Therefore, there are approximately 1.121 grams of Cl2 in the 0.500-liter sample at STP. The calculation involves converting volume to moles using the molar volume at STP and then finding the mass using the molar mass of Cl2.