how many grams of dioxygen are required to burn 5.9 grams of C2H5OH?
C2H5OH(l) +3O2(g)> 2CO2(g) +3H2O(l)
RFM of C2H5OH = (2×12)+(5×1)+16+1
=24+5+16+1
=46
No of moles= mass/RFM
No of moles of C2H5OH =5.9/46
=0.1283 moles
Mole ratio of C2H5OH to O2 is 1:3
No of moles of O2 =3/1×0.1283
=0.3849 moles
Mass= No of moles×molar mass
=0.3849×16
=6.1584g
To determine how many grams of dioxygen (O2) are required to burn 5.9 grams of C2H5OH, we need to balance the chemical equation for the combustion of ethanol (C2H5OH).
The balanced equation for the combustion of ethanol is:
C2H5OH + 3O2 -> 2CO2 + 3H2O
From the equation, we can see that for every 1 mole of C2H5OH, we need 3 moles of O2.
To convert the mass of C2H5OH to moles, we need to divide the given mass by the molar mass of C2H5OH.
The molar mass of C2H5OH = (2 * Atomic mass of C) + (6 * Atomic mass of H) + Atomic mass of O
= (2 * 12.01) + (6 * 1.01) + 16.00
= 46.07 g/mol
So, moles of C2H5OH = mass of C2H5OH / molar mass of C2H5OH
= 5.9 g / 46.07 g/mol
Now, using the balanced equation, we can calculate the moles of O2 required.
Moles of O2 = Moles of C2H5OH * (3 moles of O2 / 1 mole of C2H5OH)
Finally, we can convert the moles of O2 to grams by multiplying by the molar mass of O2, which is 32 g/mol.
Grams of O2 = Moles of O2 * Molar mass of O2
Let's calculate the answer:
Moles of C2H5OH:
5.9 g / 46.07 g/mol ≈ 0.128 moles of C2H5OH
Moles of O2:
0.128 moles of C2H5OH * (3 moles of O2 / 1 mole of C2H5OH) = 0.384 moles of O2
Grams of O2:
0.384 moles of O2 * 32 g/mol = 12.288 grams of O2
Therefore, approximately 12.288 grams of dioxygen (O2) are required to burn 5.9 grams of C2H5OH.
To determine the number of grams of dioxygen (O2) required to burn 5.9 grams of C2H5OH (ethanol), we need to first balance the chemical equation for the combustion reaction between ethanol and dioxygen.
The balanced equation for the combustion of ethanol can be written as:
C2H5OH + O2 → CO2 + H2O
From the balanced equation, we can see that one molecule of C2H5OH reacts with three molecules of O2 to produce two molecules of CO2 and three molecules of H2O.
Now, we need to calculate the molar masses of C2H5OH and O2.
The molar mass of C2H5OH can be calculated by adding up the atomic masses of all the atoms in the molecule:
Molar mass of C2H5OH = (2 * atomic mass of C) + (6 * atomic mass of H) + atomic mass of O
= (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + 16.00 g/mol
= 46.08 g/mol
The molar mass of O2 is simply the sum of the atomic masses of two oxygen atoms:
Molar mass of O2 = (2 * atomic mass of O)
= (2 * 16.00 g/mol)
= 32.00 g/mol
Now, we can proceed to calculate the amount of O2 required using stoichiometry.
Step 1: Convert the mass of C2H5OH to moles.
Number of moles of C2H5OH = (Given mass of C2H5OH) / (Molar mass of C2H5OH)
= 5.9 g / 46.08 g/mol
≈ 0.128 moles
Step 2: Use the mole ratio from the balanced equation to determine the moles of O2 required.
From the balanced equation, we see that the mole ratio between C2H5OH and O2 is 1:3.
Number of moles of O2 = (Number of moles of C2H5OH) * (Mole ratio of O2 to C2H5OH)
= 0.128 moles * (3 moles O2 / 1 mole C2H5OH)
= 0.384 moles
Step 3: Convert the moles of O2 to grams.
Mass of O2 = (Number of moles of O2) * (Molar mass of O2)
= 0.384 moles * 32.00 g/mol
= 12.29 grams (rounded to two decimal places)
Therefore, approximately 12.29 grams of dioxygen (O2) are required to burn 5.9 grams of C2H5OH.
C2H5OH + 3O2 ==> 2CO2 + 3H2O
mols C2H5OH = grams/molar mass
Convert mols C2H5OH to mols O2 using the coefficients in the balanced equation.
Now convert mols O2 to grams. g = mols x molar mass.