Sodium reacts with chlorine gas according to the reaction:\
2 Na (s) + Cl2 (g) ---> 2 NaCl (s)
What volume of chlorine gas, measured at 687 mmHg and 313 K, is required to form 275 g of NaCl?
Convert 275 g NaCl to mols.
Convert mols NaCl to mols Cl2 using the coefficients in the balanced equation.
Use PV = nRT to solve for V of Cl2 at the stated conditions.
Okay...
1) 275 g NaCl ----> 4.71 mol NaCl
2) Next, 4.71 mol NaCl * 1Cl2/2 NaCl = 2.36 mol Cl
3) V= nRT/P
v= (2.36 mol)(0.0821 L atm/K Mol)(313K)/687mmHg = 0.0883 L
Final answer: v = 0.0883 L
All of your calculations and numbers look good until the last step. Pressure in PV = nRT must go in in units of atmospheres. Change by 687/760 = ?? atm.
So, if I convert mmHg to atm I get .904 atm. When I plug that into the equation, the final answer is 67.1 L?
right
To find the volume of chlorine gas required for the reaction, we need to use the ideal gas law equation: PV = nRT.
First, let's calculate the number of moles of sodium chloride (NaCl) formed. We can do this by dividing the given mass of NaCl by its molar mass.
The molar mass of sodium chloride (NaCl) is calculated as follows:
Molar mass of Na = 22.99 g/mol
Molar mass of Cl = 35.45 g/mol
Thus, the molar mass of NaCl = (22.99 g/mol) + (35.45 g/mol) = 58.44 g/mol
Number of moles of NaCl = mass of NaCl / molar mass of NaCl = 275 g / 58.44 g/mol = 4.71 mol
According to the balanced chemical equation, the stoichiometric ratio between NaCl and Cl2 is 1:1. This means 1 mole of Cl2 is required to produce 1 mole of NaCl.
So, the number of moles of Cl2 required is also 4.71 mol.
Now we can use the ideal gas law equation, PV = nRT, to calculate the volume of chlorine gas (Cl2).
P = pressure = 687 mmHg = 0.904 atm (1 atm = 760 mmHg)
V = volume (unknown)
n = number of moles of Cl2 = 4.71 mol
R = gas constant = 0.0821 L·atm/mol·K
T = temperature = 313 K
Rearranging the equation to solve for V, we have:
V = (nRT) / P
Plugging in the values, we can calculate the volume of chlorine gas:
V = (4.71 mol * 0.0821 L·atm/mol·K * 313 K) / 0.904 atm
V ≈ 139.77 L
Therefore, the volume of chlorine gas required to form 275 g of NaCl at 687 mmHg and 313 K is approximately 139.77 liters.