How many milliliters of water will be created from a combustion reaction with 9.32×10 22nd power of ethanol molecules. assume density of water is 1.00 g/ml and molar mass of water is 18.02 g/mo

8.38 mL

Explanation:

1- The combustion reaction of ethanol (C2H5OH) in the presence of oxygen (O2) has as reaction products carbon dioxide (CO2) and water (H2O), the equation equaled is:

C2H5OH (l) + 3 O2 (g) CO2 2 CO2 (g) + 3 H2O (l)

2- By establishing the stoichiometric relationship between ethanol and water, you can calculate the number of molecules that will be created from the initial amount of alcohol molecules:

6,022x10 23 molecules of C2H5OH (1 mol) ___ 3 x 6,022x10 23 molecules of H2O (3 moles)

9.32x10 22 C2H5OH molecules X = 2.80x10 23 H2O molecules

Calculation:

9.32x10 22 x (3 x 6.022x10 23) / 6.022x10 23 = 2.80x10 23 H2O molecules

3- Once the number of water molecules formed is obtained, with the molar mass the mass can be determined:

6.022x10 23 H2O molecules 18.02 g

2.80x10 23 molecules of H2O X = 8.38 g of H2O

Calculation:

2.80x10 23 x 18.02 g / 6.022x1023 = 8.38 g of H2O

4- Finally, having the density of water, you can calculate the volume that formed:

d = m / V  --> V = m / d

V = 8.38 g / 1.00 mL = 8.38 mL

The answer is that 8.38 mL of water is formed