At STP, when 3.75 liters of oxygen reacts with excess glucose, what volume of carbon dioxide does it produce?
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44 liters
To determine the volume of carbon dioxide produced when 3.75 liters of oxygen reacts with excess glucose, we need to use the balanced chemical equation for the reaction.
The balanced equation for this reaction is:
C6H12O6 + 6O2 → 6CO2 + 6H2O
From the equation, we can see that 1 molecule of glucose reacts with 6 molecules of oxygen to produce 6 molecules of carbon dioxide.
To find the volume of carbon dioxide produced, we can use the concept of molar ratios. First, we need to convert the volume of oxygen given (3.75 liters) into moles using the ideal gas law.
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Since the question mentions "STP" (Standard Temperature and Pressure), we can assume the values of STP: P = 1 atm (atmosphere) and T = 273.15 K (Kelvin).
Using the ideal gas law, we can calculate the number of moles of oxygen:
n = PV / RT
Let's assume that the pressure is 1 atm and the temperature is 273.15 K. We can then calculate the number of moles of oxygen:
n = (1 atm) * (3.75 L) / (0.0821 L.atm/mol.K * 273.15 K) ≈ 0.172 moles of oxygen
Since the reaction is balanced in a 1:6 ratio, we can multiply the number of moles of oxygen by 6 to find the number of moles of carbon dioxide produced:
0.172 moles of O2 * 6 moles of CO2 / 1 mole of O2 = 1.032 moles of CO2
Finally, we can convert the moles of carbon dioxide back into volume using the ideal gas law:
V = nRT / P
Using the same values for pressure and temperature, and the moles we found above, we can calculate the volume of carbon dioxide produced:
V = (1.032 moles) * (0.0821 L.atm/mol.K) * (273.15 K) / (1 atm) ≈ 22.47 liters
Therefore, when 3.75 liters of oxygen reacts with excess glucose at STP, it produces approximately 22.47 liters of carbon dioxide.