What volume of chlorine gas at 34.0◦C and
1.38 atm is needed to react completely with
9.40 g of sodium to form NaCl?
Answer in units of L.
2Na + Cl2 ==> 2NaCl
mols Na = grams/atomic mass = ?
Using the coefficients in the balanced equation, convert mols Na to mols Cl2.
Then use PV = nRT to convert mols Cl2 to volume at the conditions listed.
To calculate the volume of chlorine gas needed to react with sodium and form NaCl, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure in atm
V = volume in liters (L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (K)
First, we need to find the number of moles of sodium (Na) using its molar mass. The molar mass of sodium (Na) is 22.99 g/mol.
n(Na) = Mass / Molar mass
n(Na) = 9.40 g / 22.99 g/mol
n(Na) ≈ 0.408 mol
According to the balanced chemical equation for the reaction between sodium and chlorine to form NaCl:
2 Na + Cl₂ → 2 NaCl
We can see that the ratio is 2:1 between sodium and chlorine. Therefore, the number of moles of chlorine (Cl₂) required will be half that of sodium. So:
n(Cl₂) = 0.408 mol / 2
n(Cl₂) = 0.204 mol
Now we have the number of moles of chlorine gas required. We can substitute the values into the ideal gas law equation to calculate the volume of chlorine gas.
PV = nRT
V = (nRT) / P
V = (0.204 mol * 0.0821 L·atm/(mol·K) * (34.0 + 273.15) K) / 1.38 atm
V ≈ 8.9 L
Therefore, the volume of chlorine gas needed to react completely with 9.40 g of sodium to form NaCl is approximately 8.9 L.