Glucose, c6h12o6, is metabolized in living systems. how many grams of water can be produced from the reaction of 18.0 g of glucose and 7.50 l of o2 at 1.00 atm and 37 °c? c6h12o6(s) + 6o2(g) –> 6co2(g) + 6h2o(l)

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.

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