The equation for the combustion of glucose is: C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O(g). How many grams of H2O will be produced when 8.064g of glucose is burned?

Based on the above equation, about 0.8 g of water will be produced.

To find the mass of water produced when 8.064 grams of glucose is burned, you first need to determine the moles of glucose using its molar mass, then use the stoichiometric coefficients from the balanced equation to find the moles of water produced, and finally convert the moles of water to grams using its molar mass.

1. Calculate the moles of glucose: [ \text{Moles of } C_6H_{12}O_6 = \frac{\text{mass of } C_6H_{12}O_6}{\text{molar mass of } C_6H_{12}O_6} ]

2. Use stoichiometry to find moles of water produced: [ \text{Moles of } H_2O = \text{Moles of } C_6H_{12}O_6 \times \frac{6 \text{ moles } H_2O}{1 \text{ mole } C_6H_{12}O_6} ]

3. Convert moles of water to grams: [ \text{Mass of } H_2O = \text{Moles of } H_2O \times \text{molar mass of } H_2O ]

Perform the calculations:

1. Moles of glucose: [ \text{Moles of } C_6H_{12}O_6 = \frac{8.064 \text{ g}}{180.16 \text{ g/mol}} = 0.0447 \text{ mol} ]

2. Moles of water produced: [ \text{Moles of } H_2O = 0.0447 \text{ mol} \times \frac{6 \text{ mol } H_2O}{1 \text{ mol } C_6H_{12}O_6} = 0.268 \text{ mol} ]

3. Mass of water produced: [ \text{Mass of } H_2O = 0.268 \text{ mol} \times 18.015 \text{ g/mol} = 4.83 \text{ g} ]

Therefore, when 8.064 grams of glucose is burned, 4.83 grams of water will be produced.