How much carbon dioxide measured at 25.0◦C
and 821 Torr would be produced by the combustion
of 319 g of glucose (C6H12O6)?
1. 40.1 L
2. 20.2 L
3. 241 L
4. 3.37 L
Anna's answer is wrong.
241 L is the correct answer.
bruh anna how u this slow
Try PV=nRT, where R=0.08206, P=pressure, V=volume, n=moles, and T=temperature. Convert temperature to Kelvin, Torr to atm (pressure), and g to moles (divide by molar mass).
the question asked how much CO2 measured.
thus, the answer is 241L
Anna's answer is the volume of glucose(C6H12O6)
anna you made me fail chemistry and i got rejected from community college why would you do this to me
I'm sorry to hear that. It's always important to double check your work and make sure your answers are correct. It's also important to seek help from your instructor or tutor if you're struggling with a subject. Don't give up and keep working hard!
To find the amount of carbon dioxide produced by the combustion of glucose, we need to use the stoichiometry of the balanced chemical equation for the combustion reaction. The balanced equation for the combustion of glucose is:
C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O
From the equation, we can see that 1 mole of glucose (C6H12O6) reacts to produce 6 moles of carbon dioxide (CO2).
First, let's calculate the number of moles of glucose present in 319 g of glucose. The molar mass of glucose is 180.16 g/mol.
Number of moles of glucose = mass of glucose / molar mass of glucose
= 319 g / 180.16 g/mol
= 1.7708 mol (rounded to four decimal places)
Now, we can calculate the number of moles of carbon dioxide produced.
Number of moles of carbon dioxide = moles of glucose × (6 moles of CO2 / 1 mole of glucose)
= 1.7708 mol × 6
= 10.6248 mol (rounded to four decimal places)
Next, we can use the ideal gas law to convert the number of moles of carbon dioxide to volume. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in atm or Torr)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
We have the pressure (821 Torr) and the temperature (25.0 °C). To convert °C to Kelvin, we add 273.15.
Temperature in Kelvin = 25.0 °C + 273.15
= 298.15 K
Now we can calculate the volume of carbon dioxide.
V = nRT / P
= (10.6248 mol) × (0.0821 L·atm/mol·K) × (298.15 K) / (821 Torr)
= 0.3949 L (rounded to four decimal places)
Therefore, the answer is approximately 0.395 L. Out of the options given, the closest value is 0.4 L or 3.37 L.
The correct answer would be 4. 3.37 L.
Anna is wrong and should probably drop out for her stupidity. The real answer is 241 L
Answer is 40.1L.
After conversions, plug it all into V=nRT/P