Respuesta :
Answer:
1. 20.5 L of CO₂ are released.
2. 44.8L of O₂ are needed
Explanation:
C₆H₁₂O₆(aq) + 6O₂(g) → 6CO₂ (g) + 6H₂O(l)
We assume oxygen as reagent in excess.
Let's convert the mass of glucose to moles → 24 g . 1mol/180 g = 0.133 moles
Ratio is 1:6 so (0.133 . 6) = 0.8 moles of CO₂ are released
We apply the Ideal Gases Law to determine the volume:
P . V = n . R .T → V = (n. R .T)/P
V = (0.8 mol . 0.082 . 310K) /0.990 atm = 20.5L
For 2nd question: First of all, we determine the moles of glucose
55 g . 1mol/180 g = 0.305 moles
We apply a rule of three: 1 mol of glucose needs 6 moles of O₂ to react
0.305 moles may need (0.305 .6)/1 = 1.83 moles
We use the Ideal Gases Law again:
V = (1.83 mol . 0.082 . 298K) / 1 atm = 44.8L
Answer:
1) 20.5 L CO2
2) 44.75 L O2
Explanation:
Step 1: Data given
Temperature = 37.0 °C
Pressure = 0.990 atm
Mass of glucose = 24.0 grams
Molar mass glucose: 180.156 g/mol
Step 2: The balanced equation
C6H12O6(aq)+6O2(g) → 6CO2(g) + 6H2O(l)
Step 3: Calculate moles glucose
Moles glucose = mass glucose / molar mass glucose
Moles glucose = 24.0 grams / 180.156 g/mol
Moles glucose = 0.133 moles
Step 4: Calculate moles CO2
For 1 mol glucose we need6 moles O2 to produce 6 moles CO2 and 6 moles H2O
For 0.133 moles glucose we'll have 6*0.133 = 0.798 moles CO2
Step 5: Calculate volume CO2
p*V = n*R*T
⇒ p = the pressure = 0.990 atm
⇒ V = the volume of CO2 = TO BE DETERMINED
⇒ n = the number of moles CO2 = 0.798 moles
⇒ R = the gas constant = 0.08206 L* atm / K*mol
⇒ T = the temperature = 37.0 °C = 310 K
V = (nRT)/p
V = (0.798 * 0.08206 * 310) / 0.990
V= 20.5 L CO2
2.) Calculate the volume of oxygen you would need, at 1.00 atm and 298 K, to completely oxidize 55g of glucose.
Calculate moles glucose
Moles glucose = mass glucose / molar mass glucose
Moles glucose = 55.0 grams / 180.156 g/mol
Moles glucose = 0.305 moles
Calculate moles O2
For 1 mol glucose we need6 moles O2 to produce 6 moles CO2 and 6 moles H2O
For 0.305 moles we need 6 * 0.305 = 1.83 moles O2
Calculate volume of O2
p*V = n*R*T
⇒ p = the pressure = 1.00 atm
⇒ V = the volume of O2 = TO BE DETERMINED
⇒ n = the number of moles O2 = 1.83 moles
⇒ R = the gas constant = 0.08206 L* atm / K*mol
⇒ T = the temperature = 298 K
V = (nRT)/p
V = (1.83 * 0.08206 * 298) / 1.00
V= 44.75 L O2
