Consider the reaction:
CO(g)+Cl2(g)⇌COCl2(g)
Keq= 2.9×1010 at 25 ∘C
A 5.89 −L flask containing an equilibrium reaction mixture has [CO]= 1.9×10−6 M and [Cl2]= 9.3×10−7 M .
How much COCl2 in grams is in the equilibrium mixture?
Keq expression for the following: Co + Cl2↔CoCl2
To find the amount of COCl2 in grams in the equilibrium mixture, you'll need to use the given concentrations of CO and Cl2 and the volume of the flask to calculate the moles of COCl2. Then, you can convert the moles of COCl2 to grams using the molar mass of COCl2.
1. Calculate the moles of COCl2:
Since the given concentrations are in Molarity (M), you can use the formula:
Moles = Concentration (M) x Volume (L)
Moles of COCl2 = [COCl2] x Volume of flask (L)
Moles of COCl2 = (1.9x10^-6 M) x (5.89 L)
2. Use the balanced equation to determine the stoichiometry:
The balanced equation for the reaction is: CO(g) + Cl2(g) ⇌ COCl2(g)
According to the balanced equation, the mole ratio of CO to COCl2 is 1:1.
Since the moles of CO and COCl2 are equal, the moles of COCl2 are also given by:
Moles of COCl2 = Moles of CO
3. Convert moles of COCl2 to grams:
The molar mass of COCl2 can be obtained from the periodic table by summing the atomic masses of carbon, oxygen, and chlorine.
Molar mass of COCl2 = (12.01 g/mol) + (16.00 g/mol) + (35.45 g/mol) + (35.45 g/mol)
Finally, you can calculate the mass of COCl2:
Mass of COCl2 = Moles of COCl2 x Molar mass of COCl2
Now, let's plug in the values and calculate:
Moles of COCl2 = (1.9x10^-6 M) x (5.89 L) = 1.12x10^-8 moles
Molar mass of COCl2 = (12.01 g/mol) + (16.00 g/mol) + (35.45 g/mol) + (35.45 g/mol) = 98.91 g/mol
Mass of COCl2 = (1.12x10^-8 moles) x (98.91 g/mol) = 1.11x10^-6 grams
Therefore, there is approximately 1.11x10^-6 grams of COCl2 in the equilibrium mixture.
........CO(g) + Cl2(g)⇌ COCl2(g)
E.....1.9E-6...9.3E-7.....x
Substitute the E line into the Keq expression and solve for x = (COCL2) in mols/L.
Then mols/L x 5.89 L = mols COCl2.
Grams COCl2 = mols COCl2 x molar mass COCl2.