Could someone walk me through this problem?
6CO2 + 6H2O --> C6H12O6 + 6O2
partial pressure of CO2 is .26Torr, temp is 25 celsius, calculate volume of air to prouduce 10g glucose
I keep getting confused with stoichiometry....my thinking: convert 10 g glucose to moles (which I found to be .0556 mol) and then I'll need that because there's a 6:1 mole relationship with CO2..but I'm not completely sure how to use this.
moles glucose = 10g/180.16 = ??
moles CO2 = moles glucose x (6 moles CO2/1 mole glucose) = 6 x moles glucose.
To solve this problem, you are on the right track with using stoichiometry. Let's break it down step by step:
Step 1: Convert 10 g of glucose to moles. To do this, we need to know the molar mass of glucose. The molar mass of glucose (C6H12O6) is approximately 180.16 g/mol. Therefore, we can calculate the number of moles of glucose as follows:
moles of glucose = mass of glucose / molar mass
moles of glucose = 10 g / 180.16 g/mol ≈ 0.0555 mol
Step 2: Determine the mole ratio between glucose and carbon dioxide (CO2) using the balanced equation. From the equation, we can see that for every 1 mole of glucose produced, 6 moles of carbon dioxide are consumed.
Therefore, the mole ratio of glucose to carbon dioxide is 1:6. This means that for the production of 0.0555 moles of glucose, we need 0.0555 * 6 = 0.333 moles of carbon dioxide.
Step 3: Convert moles of carbon dioxide to volume using the ideal gas law. The ideal gas law equation is as follows:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L*atm/(mol*K)), and T is the temperature in Kelvin.
First, we need to convert the temperature from Celsius to Kelvin by adding 273.15:
T = 25°C + 273.15 = 298.15 K
Now, we can solve for volume. Rearranging the ideal gas law equation, we get:
V = nRT / P
where V is the volume, n is the number of moles of gas, R is the ideal gas constant, and P is the pressure.
V = 0.333 mol * 0.0821 L*atm/(mol*K) * 298.15 K / 0.26 atm ≈ 32.7 L
So, the volume of air required to produce 10 g of glucose at a partial pressure of CO2 of 0.26 Torr and a temperature of 25°C is approximately 32.7 liters.