The density of ethanol, C2H5OH, is 0.789 g/mL. How many milliliters of ethanol are needed to produce 30.0 g of CO2 according to the following chemical equation?
C2H5OH(I) + 3 O2(g) → 2 CO2(g) + 3H20(I)
To solve this problem, we first need to calculate the molar mass of ethanol (C2H5OH) and CO2 to determine the stoichiometry of the reaction.
Molar Mass of C2H5OH:
C = 12.01 g/mol
H = 1.008 g/mol (5 hydrogen atoms)
O = 16.00 g/mol
Molar mass of C2H5OH = 12.01 + (1.008 x 5) + 16.00 = 46.08 g/mol
Now, we need to calculate the number of moles of C2H5OH that will produce 30.0 g of CO2 using the stoichiometry of the reaction.
The balanced equation shows that 1 mol of C2H5OH reacts to produce 2 mol of CO2. Therefore, the molar ratio is 1:2.
Number of moles of C2H5OH = 30.0 g / 46.08 g/mol = 0.650 mol
From the stoichiometry, we know that 1 mol of C2H5OH produces 2 mol of CO2.
Number of moles of CO2 produced = 0.650 mol x 2 = 1.30 mol
Finally, we can calculate the volume of ethanol needed using the density:
Volume of ethanol = 1.30 mol x 46.08 g/mol / 0.789 g/mL = 75.84 mL
Therefore, 75.84 mL of ethanol are needed to produce 30.0 g of CO2.