How many grams of O2 are released into the environment during the production of the 5.1 grams of glucose?
To determine the amount of oxygen released during glucose production, we need to consider the chemical equation for cellular respiration. Glucose (C6H12O6) reacts with oxygen (O2) and produces carbon dioxide (CO2), water (H2O), and energy.
The balanced equation for this reaction is:
C6H12O6 + 6O2 → 6CO2 + 6H2O + Energy
From the equation, we can see that for every 1 mole of glucose consumed, 6 moles of oxygen are required.
To find the amount of oxygen released for the given mass of glucose, we need to calculate the number of moles of glucose and then use the mole ratio from the balanced equation.
The molar mass of glucose (C6H12O6) is calculated as follows:
6 carbon atoms * atomic mass of carbon (12.01 g/mol) = 72.06 g/mol
12 hydrogen atoms * atomic mass of hydrogen (1.01 g/mol) = 12.12 g/mol
6 oxygen atoms * atomic mass of oxygen (16.00 g/mol) = 96.00 g/mol
Total molar mass of glucose = 72.06 + 12.12 + 96.00 = 180.18 g/mol
Next, we can calculate the moles of glucose by dividing the given mass by the molar mass:
moles of glucose = mass of glucose / molar mass of glucose
moles of glucose = 5.1 g / 180.18 g/mol = 0.028 moles (rounded to 3 decimal places)
Now, using the mole ratio from the balanced equation, we know that 1 mole of glucose reacts with 6 moles of oxygen.
moles of oxygen = moles of glucose * (moles of oxygen / moles of glucose)
moles of oxygen = 0.028 moles * (6 moles of oxygen / 1 mole of glucose) = 0.168 moles (rounded to 3 decimal places)
Finally, to convert moles of oxygen to grams, we multiply by the molar mass of oxygen:
mass of oxygen = moles of oxygen * molar mass of oxygen
mass of oxygen = 0.168 moles * 32.00 g/mol = 5.376 g (rounded to 3 decimal places)
Therefore, approximately 5.376 grams of O2 are released into the environment during the production of 5.1 grams of glucose.