What is the percent yield of CO2 if a reaction using 10.0 g CO with excess O2 produces 12.8 g CO2?
2CO + O2 → 2CO2
I think its 78%
Well, let's do some math and find out!
To find the percent yield, you first need to calculate the theoretical yield, which is the maximum amount of product that could be formed based on the given reactants. In this case, the balanced equation tells us that 2 moles of CO react with 1 mole of O2 to produce 2 moles of CO2.
So, let's convert the given 10.0 g of CO to moles. Using the molar mass of CO (28.01 g/mol), we can determine that 10.0 g of CO is equivalent to 0.357 moles of CO.
Since the molar ratio in the balanced equation is 2:2, we know that 0.357 moles of CO will produce 0.357 moles of CO2.
Now, let's convert moles of CO2 to grams. The molar mass of CO2 is 44.01 g/mol, so 0.357 moles of CO2 is equal to 15.6 g.
Therefore, the theoretical yield of CO2 is 15.6 g.
To calculate the percent yield, you need to divide the actual yield (12.8 g) by the theoretical yield (15.6 g) and then multiply by 100.
(12.8 g / 15.6 g) * 100 = 82.05%
So, the percent yield of CO2 in this reaction is approximately 82.05%. Keep in mind that any discrepancy between the actual and theoretical yields could be due to factors such as experimental error, side reactions, or incomplete conversions.
I hope that wasn't too theoretical for you! Remember, always yield to the math.
To calculate the percent yield of CO2, we need to compare the actual yield (which is given as 12.8 g) to the theoretical yield.
Step 1: Determine the molar masses of CO and CO2.
- Molar mass of CO: 12.01 g/mol (C) + 16.00 g/mol (O) = 28.01 g/mol
- Molar mass of CO2: 12.01 g/mol (C) + (2 * 16.00 g/mol (O)) = 44.01 g/mol
Step 2: Calculate the moles of CO used.
- Moles of CO = Mass / Molar mass = 10.0 g / 28.01 g/mol = 0.356 mol CO
Step 3: Use stoichiometry to determine the theoretical yield of CO2.
From the balanced chemical equation, we can see that 2 moles of CO react to produce 2 moles of CO2.
- Moles of CO2 = (2/2) * Moles of CO = 0.356 mol CO2
Step 4: Calculate the theoretical yield of CO2 in grams.
- Theoretical yield of CO2 = Moles of CO2 * Molar mass of CO2 = 0.356 mol * 44.01 g/mol = 15.65 g CO2
Step 5: Calculate the percent yield of CO2.
- Percent yield = (Actual yield / Theoretical yield) * 100
- Percent yield = (12.8 g / 15.65 g) * 100 = 81.78%
Therefore, the percent yield of CO2 in this reaction is approximately 81.78%.
To calculate the percent yield of CO2 in this reaction, we need to know the theoretical yield and the actual yield.
The theoretical yield is the amount of product that would be obtained if the reaction proceeded perfectly, assuming all reactants were completely consumed. The actual yield, on the other hand, is the amount of product obtained in a real experiment.
In this case, we are given the actual yield of CO2, which is 12.8 g. To find the theoretical yield, we need to determine the limiting reactant in the reaction.
To find the limiting reactant, we compare the moles of each reactant to the stoichiometric ratio in the balanced chemical equation. In this case, we have:
10.0 g CO (Convert to moles)
12.8 g CO2 (Convert to moles)
The molar mass of CO is 28.01 g/mol, and the molar mass of CO2 is 44.01 g/mol.
10.0 g CO × (1 mol CO / 28.01 g CO) = 0.357 mol CO
12.8 g CO2 × (1 mol CO2 / 44.01 g CO2) = 0.291 mol CO2
From the balanced chemical equation (2CO + O2 → 2CO2), we can see that the stoichiometric ratio between CO and CO2 is 2:2 or 1:1.
Since the ratio of CO to CO2 in the reaction is 1:1, the limiting reactant is CO. This means that CO is completely consumed, and any excess O2 remains unreacted.
Since we now know that CO is the limiting reactant and that the stoichiometric ratio between CO and CO2 is 1:1, the theoretical yield of CO2 is also 0.357 mol.
Next, we need to convert the theoretical yield from moles to grams using the molar mass of CO2:
0.357 mol CO2 × (44.01 g CO2 / 1 mol CO2) = 15.7 g CO2
Now, to calculate the percent yield, we use the formula:
Percent yield = (actual yield / theoretical yield) × 100
Percent yield = (12.8 g / 15.7 g) × 100 = 81.5%
Therefore, the percent yield of CO2 in this reaction is 81.5%.
molar mass of CO2 = 12 + 2*16 = 44 grams / mol
molar mass of CO = 12 + 16 = 28 grams / mol
10 g CO = 10/28 = 0.357 mols CO in
if perfect should then get 0.357 mols CO2 out
12.2 g CO2 = 12.2/44 = 0.277 mols CO2 produced
Oh dear, got less out than perfection demands
so got .277/.357 = about 78 % of perfection