Answer:
Percent error = [(1600 g - 58.608 g) / 58.608 g] x 100%
Percent error = 2640.02%
Step-by-step explanation:
To calculate the percent error of this reaction, we need to first calculate the theoretical yield of CO2 based on the balanced equation and then compare it to the actual yield obtained.
From the balanced equation:
C6H12O6 + 6O2 → 6CO2 + 6H2O
We can see that 1 mole of glucose (C6H12O6) produces 6 moles of carbon dioxide (CO2). The molar mass of glucose is 180 g/mol. So, to find the theoretical yield of CO2, we can use the following steps:
---Convert the mass of glucose to moles:
40 g / 180 g/mol = 0.222 mol
---Calculate the moles of CO2 produced:
0.222 mol glucose x 6 mol CO2/mol glucose = 1.332 mol CO2
---Convert the moles of CO2 to grams:
1.332 mol x 44 g/mol = 58.608 g CO2
So, the theoretical yield of CO2 is 58.608 grams.
Now we can calculate the percent error using the following formula:
Percent error = [(experimental value - theoretical value) / theoretical value] x 100%
Plugging in the values we have:
Percent error = [(1600 g - 58.608 g) / 58.608 g] x 100%
Percent error = 2640.02%
This means that the experimental value is significantly higher than the theoretical value, which indicates a large error in the experiment. It's important to identify and correct sources of error in experiments to improve the accuracy of results.