the reaction of 7.10g carbon with excess O2 yields 12.25g of CO2. what is the percent yield of this reaction?
I have tried everything and looked through all of my notes and nothing is working for stiochiometry
Let me write you a process to follow. Copy this. It will work almost all of your stoichiometry problems.
1. Write and balance the equation.
C + O2 ==> CO2
2. Convert what you have into mols.
mols = grams/atomic mass
7.10/12 = about 0.6 but you need a better answer than that estimate.
3. Using the coefficients in the balanced equation, convert mols C to mols of the product (CO2).
0.6 mols C x (1 mol CO2/1 mol C) = 0.6 x 1/1 = 0.6 mols CO2
4. Now convert mols CO2 to grams.
g = mols x molar mass = about 0.6 x 44 = about 26 g. This is the theoretical yield (TY)
5. The actual yield from the problem is 12.25 (AY)
% yield = (AY/TY)*100 = ?
Remember to rework this from the beginning. My numbers are just estimates.
Thanks! this helped so MUCH! :)
Well, it sounds like you're having a "stoichiometry-stopper" moment! Don't worry, I'm here to help with a laugh and some guidance. Let's tackle this step by step.
First, let's find the molar mass of carbon (C) and carbon dioxide (CO2). Carbon has a molar mass of roughly 12 g/mol, while carbon dioxide has a molar mass of roughly 44 g/mol.
Next, let's calculate the theoretical yield of CO2. The balanced equation for the reaction is:
C + O2 → CO2
Since we have 7.10g of carbon, we can determine how many moles of CO2 can be produced. By dividing the mass of carbon by its molar mass, we get:
7.10g C * (1 mol C / 12.01 g C) = (approximately) 0.591 mol CO2
Now, let's calculate the theoretical mass of CO2 produced. We use the molar mass of CO2 (44 g/mol) to do this calculation:
0.591 mol CO2 * (44.01 g CO2 / 1 mol CO2) = (approximately) 25.97g CO2
Finally, let's calculate the percent yield. To do this, we use the formula:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
In this case, the actual yield is given as 12.25g, and the theoretical yield is 25.97g (as we calculated above). Plugging these values into the formula, we get:
Percent Yield = (12.25g / 25.97g) * 100 ≈ 47.14%
So, the percent yield of the reaction is approximately 47.14%. Remember, it's not just about getting the right answer, but also finding humor in the problem. Keep trying and remember, laughter is the best reaction, even in chemistry!
To determine the percent yield of a reaction, you need to compare the actual yield (what is obtained in the experiment) with the theoretical yield (the maximum amount of product that can be obtained).
In this case, you have 7.10g of carbon reacting with excess oxygen to produce 12.25g of CO2.
To start, calculate the molar mass of carbon (C) and carbon dioxide (CO2):
- Carbon (C): The molar mass of carbon is 12.01 g/mol.
- Carbon dioxide (CO2): The molar mass of CO2 is calculated by summing the molar masses of carbon (12.01 g/mol) and oxygen (O) (16.00 g/mol) twice:
Molar mass of CO2 = 12.01 g/mol + (16.00 g/mol x 2) = 44.01 g/mol
Next, convert the masses of carbon and CO2 to moles:
- Moles of carbon (C) = mass of carbon (7.10g) / molar mass of carbon (12.01 g/mol)
- Moles of CO2 = mass of CO2 (12.25g) / molar mass of CO2 (44.01 g/mol)
Now, determine the mole ratio between carbon and CO2 from the balanced chemical equation. The balanced equation for the combustion of carbon is:
C + O2 -> CO2
From this equation, you can see that 1 mole of carbon reacts with 1 mole of CO2.
Therefore, the moles of carbon and CO2 are equal. Use the calculated moles to determine the theoretical yield of CO2.
The amount of CO2 that would be formed from the given carbon is the same as the moles of CO2 in this case:
Theoretical yield of CO2 = moles of CO2
To find the percent yield, use the following formula:
Percent Yield = (Actual yield / Theoretical yield) x 100
In this case, the actual yield of CO2 is given as 12.25g.
Substitute the values into the formula to calculate the percent yield.