A reaction between liquid reactants takes place at - 16C in a sealed, evacuated vessel with a measured volume of 45L. Measurements show that the reaction produced 34g of sulfur hexafluoride gas. Calculate the pressure of sulfur hexafluoride gas in the reaction vessel after the reaction. You may ignore the volume of the liquid reactants. Be sure your answer has the correct number of significant digits.

0.11 atm

Explanation:

Given data

Volume (V) : 45 L Temperature (T) : - 16 °C + 273.15 = 257 K Mass (m) : 34 g Ideal gas constant (R) : 0.082 atm. L/mol. K

First, we will calculate the moles (n) of sulfur hexafluoride, considering its molar mass is 146.06 g/mol.

34 g * (1 mol/146.06 g) = 0.23 mol

Then, we can calculate the pressure (P) of sulfur hexafluoride using the ideal gas equation.

P * V = n * R * T

P = n * R * T / V

P = 0.23 mol * (0.082 atm. L/mol. K) * 257 K / 45 L

P = 0.11 atm