Answer:
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:
<em>C2H5OH (l) + 3 O2 (g) CO2 2 CO2 (g) + 3 H2O (l)
</em>
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
<em>Calculation:
</em>
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
<em>Calculation:
</em>
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
