When NO(g) (0.06623 mol/L) and 422.6 grams of Cl2(g) in a 180.0 L reaction vessel at 565.0 °C are allowed to come to equilibrium the mixture contains 3.096 mol of NOCl(g). What is the concentration (mol/L) of NOCl(g)?
2NO(g)+Cl2(g) = 2NOCl(g)
To find the concentration (mol/L) of NOCl(g), we need to calculate the number of moles of NOCl(g) and then divide it by the volume of the reaction vessel.
First, let's calculate the number of moles of NOCl(g) using the given information. We are told that the initial concentration of NO(g) is 0.06623 mol/L and the volume of the reaction vessel is 180.0 L. We also know that the initial number of moles of NOCl(g) is given as 3.096 mol.
According to the balanced chemical equation, 2 moles of NO(g) react with 1 mole of Cl2(g) to form 2 moles of NOCl(g). This means that the number of moles of NOCl(g) formed is half the number of moles of NO(g) that react.
Therefore, the number of moles of NO(g) that react is 3.096 mol/2 = 1.548 mol.
Since the initial concentration of NO(g) is given as 0.06623 mol/L, we can calculate the initial number of moles of NO(g) using the equation:
number of moles = concentration x volume
number of moles of NO(g) = 0.06623 mol/L x 180.0 L = 11.814 mol
Now we can calculate the number of moles of NOCl(g) that are formed by subtracting the number of moles of NO(g) that react from the initial number of moles of NOCl(g):
number of moles of NOCl(g) formed = initial number of moles of NOCl(g) - number of moles of NO(g) that react
= 3.096 mol - 1.548 mol
= 1.548 mol
Finally, we can calculate the concentration of NOCl(g) by dividing the number of moles of NOCl(g) formed by the volume of the reaction vessel:
concentration of NOCl(g) = moles of NOCl(g) formed / volume
concentration of NOCl(g) = 1.548 mol / 180.0 L
Therefore, the concentration of NOCl(g) is approximately 0.0091 mol/L.