Respuesta :
Answer:
There is 5.32 grams of water produced
Explanation:
Step 1: Data given
Mass of glucose = 18.0 grams
Molar mass of glucose = 180.156 g/mol
Volume of O2 = 7.50 L
Temperature = 37 °C
Step 2: The balanced equation
C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l)
Step 3: Calculate moles of glucose
Moles glucose = mass glucose / molar mass glucose
Moles glucose = 18.0 grams / 180.156 g/mol
Moles glucose = 0.0999 moles
Step 4: Calculate moles of O2
At STP, 1 mol of gas = 22.4 L. At 37 °C (= 310 K) and 1 atm, 1 mol = (22.4 L)(310 K/273 K) = 25.4 L.
Therefore the amount of O₂ here is (7.50 L) / (25.4 L/mol) = 0.295 moles
Step 5: Determine the limiting reactant
O2 is the limiting reactant. It will completely be consumed (0.295 moles).
Glucose is in excess. There will react 0.295/6 = 0.0492 moles
There will remain 0.0999 - 0.0492 = 0.0507 moles
Step 6: Calculate moles of water
For 1 mol glucose, we need 6 moles of O2 to produce 6 moles of CO2 and 6 moles of H2O
For 0.295 moles of O2, we'll have 0.295 moles of H2O
Step 7: Calculate mass of water
Mass of water = moles water * molar mass water
Mass of water = 0.295 mol * 18.02 g/mol
Mass of water = 5.32 grams
There is 5.32 grams of water produced
Note that the C₆H₁₂O₆ and the O₂ react in a 6:1 ratio. The molar mass of C₆H₁₂O₆ is 180.16 g/mol, so there are (24.5 g)/(180.16 g/mol) = 0.136 mol here. This can react with 6(0.136) = 0.816 mol O₂.
Since there are just 0.248 mol O₂, oxygen is the limiting reactant. (1/6)(0.248 mol) = 0.0413 mol glucose is consumed.
The water is produced in a 6:1 ratio with the glucose consumed, so 0.248 mol are produced.
(0.248 mol)(18.02 g/mol) = 4.47 g H₂O.
