What volume of Cl2 gas, measured at 687torr and 34∘C, is required to form 27g of NaCl?

Cl2(gas) ==> 2NaCl which isn't balanced but the Cl is balanced and that's what the question is about.

mols NaCl = grams/molar mass NaCl
Using the coefficients in the balanced equation above to convert mols NaCl to mols Cl2. Then use PV = nRT and solve for V Cl2 in L.

Thank you

To find the volume of Cl2 gas required to form 27g of NaCl, we need to use the stoichiometry of the balanced chemical equation.

The balanced chemical equation for the reaction between Cl2 gas and Na is:

2 Na + Cl2 -> 2 NaCl

From the equation, we can see that 2 moles of Na reacts with 1 mole of Cl2 to form 2 moles of NaCl.

Step 1: Calculate the moles of NaCl
Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol

Moles of NaCl = mass / molar mass
Moles of NaCl = 27g / 58.44 g/mol = 0.462 mol

Step 2: Calculate the moles of Cl2 required
From the stoichiometry of the balanced equation, we know that 2 moles of NaCl reacts with 1 mole of Cl2.

Moles of Cl2 = 0.462 mol / 2 = 0.231 mol

Step 3: Convert moles to volume using the ideal gas law

The ideal gas law is given by:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (in Kelvin)

We need to convert the given temperature from Celsius to Kelvin:

T(K) = T(°C) + 273.15
T(K) = 34 + 273.15 = 307.15 K

Now we can plug in the values into the ideal gas law equation:

PV = nRT
V = (nRT) / P

V = (0.231 mol * 0.0821 L.atm/mol.K * 307.15 K) / 687 torr

Step 4: Convert the pressure from torr to atm:
1 atm = 760 torr

Pressure (P) = 687 torr / 760 torr/atm = 0.903 atm

V = (0.231 mol * 0.0821 L.atm/mol.K * 307.15 K) / 0.903 atm
V ≈ 5.96 L

Therefore, approximately 5.96 liters of Cl2 gas at 687 torr and 34°C is required to form 27g of NaCl.

To determine the volume of Cl2 gas required to form 27g of NaCl, we need to use the stoichiometric ratio between Cl2 and NaCl. The balanced chemical equation for the reaction is:

2 Na + Cl2 -> 2 NaCl

First, we need to convert the given mass of NaCl (27g) to moles. The molar mass of NaCl is 22.99 g/mol for Na and 35.45 g/mol for Cl.

Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol

Moles of NaCl = mass / molar mass = 27g / 58.44 g/mol ≈ 0.463 mol

According to the balanced chemical equation, 1 mole of Cl2 reacts with 2 moles of NaCl. So, the moles of Cl2 required can be calculated using the stoichiometric ratio:

Moles of Cl2 = (moles of NaCl) / (2 moles of NaCl per mole of Cl2) = 0.463 mol / 2 ≈ 0.232 mol

Now we can apply the ideal gas law to calculate the volume of Cl2 gas. The ideal gas law equation is:

PV = nRT

Where:
P = pressure in atm
V = volume in liters
n = moles
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin

In this case, we are given:
P = 687 torr (convert to atm by dividing by 760 torr/atm)
T = 34°C (convert to Kelvin by adding 273.15)

P = 687 torr / 760 torr/atm = 0.903 atm
T = 34°C + 273.15 = 307.15 K

Now, we can rearrange the ideal gas law equation to solve for volume:

V = (nRT) / P

Plugging in the values:

V = (0.232 mol)(0.0821 L.atm/mol.K)(307.15K) / 0.903 atm ≈ 5.96 L

Therefore, approximately 5.96 liters of Cl2 gas at a pressure of 687 torr and a temperature of 34°C are required to form 27g of NaCl.