ClO2 is sometimes used as a chlorinating agent for water treatment.
Cl2 + 4 NaClO -> 4 NaCl + 2 ClO2
In an experiment, 1L Cl2 measured at 10C and 4.66atm is dissolved in .75L of 2.00M NaClO. If 25.9g of pure ClO2 is obtained, then what is the percent yeild for this experiment?
This is what I got:
I calculated the moles of Cl2 using PV=nrt, and got n= .200. For mass, I got Cl2= 14.18g
Then to get the moles of ClO2, I did a ration, .200/x = 1/2, x= .40 moles ClO2
Mass ClO2 = .4 X 67.45g/mol = 26.98g
Percent yield= 14.18g/26.98g x 100 = 52.56
Please tell me if there are any mistakes. Thank you!
Everything is ok except for the last step. The question is for % ClO2 and you substituted mass Cl2 and not ClO2.
That should be (25.9/26.98)*100 = ? and round to 3 significant figures.
Here is a short cut to calculating mols ClO2. You used ratio/proportion and that will work every time but you can also use the numbers in the balanced equation.
0.2 mol Cl2 x (2 mols ClO2/1 mol Cl2) = 0.4 mol ClO2. That's easy enough to do in your head whereas if I set it up with ratio/proportion it is harder to do in my head.
Thanks for showing your work. It helps us help you.
Your calculations seem to be correct based on the information provided. Here's a step-by-step breakdown of the calculations:
1. Calculate the moles of Cl2:
Using the ideal gas law equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
P = 4.66 atm
V = 1 L
T = 10 + 273.15 (convert temperature from Celsius to Kelvin)
R = 0.0821(atm•L)/(mol•K)
Using PV = nRT, we can rearrange the equation to solve for n:
n = PV / RT
= (4.66 atm) * (1 L) / (0.0821 (atm•L)/(mol•K) * (10 + 273.15) K)
= 0.1998 mol
2. Calculate the mass of Cl2:
The molar mass of Cl2 is 35.45 g/mol.
Mass of Cl2 = 0.1998 mol * 35.45 g/mol = 7.09 g
3. Calculate the number of moles of ClO2 using the stoichiometric ratio:
From the balanced equation, 1 mole of Cl2 produces 2 moles of ClO2.
So, the moles of ClO2 = (0.1998 mol Cl2) * (2 mol ClO2 / 1 mol Cl2) = 0.3996 mol ClO2
4. Calculate the mass of ClO2:
The molar mass of ClO2 is 67.45 g/mol.
Mass of ClO2 = 0.3996 mol * 67.45 g/mol = 26.94 g (approximately)
5. Calculate the percent yield:
The percent yield is calculated by dividing the actual yield (mass of ClO2 obtained) by the theoretical yield (mass of ClO2 calculated) and multiplying by 100.
Percent yield = (actual yield / theoretical yield) * 100
= (25.9 g / 26.94 g) * 100
≈ 96.1%
Therefore, the percent yield for this experiment is approximately 96.1%.
To calculate the percent yield, you first need to determine the theoretical yield, which is the maximum amount of product that can be obtained from the given reactants.
From the balanced equation:
1 mole of Cl2 produces 2 moles of ClO2
So, if you have 0.2 moles of Cl2, the theoretical yield of ClO2 would be:
0.2 moles Cl2 × (2 moles ClO2 / 1 mole Cl2) = 0.4 moles ClO2
Now, to calculate the mass of ClO2 that would be obtained from the theoretical yield, you can use the molar mass of ClO2:
Mass ClO2 = 0.4 moles ClO2 × 67.45 g/mol = 26.98 g
So, your calculation for the mass of ClO2 is correct.
Finally, to calculate the percent yield, you need to divide the actual yield (25.9 g) by the theoretical yield (26.98 g), and then multiply by 100:
Percent yield = (25.9 g / 26.98 g) × 100 = 95.94%
Therefore, the correct percent yield for this experiment is approximately 95.94%, not 52.56%.