how many liters of oxygen are necessary for the combustion of 340 g of ethanol (c2h5OH) assuming the raction occurs at STP? no idea how to work this out
To calculate the number of liters of oxygen necessary for the combustion of ethanol (C2H5OH), we need to use stoichiometry and the balanced chemical equation for the combustion reaction.
The balanced chemical equation for the combustion of ethanol is:
C2H5OH + 3O2 → 2CO2 + 3H2O
From the balanced equation, we can see that one mole of ethanol reacts with three moles of oxygen to produce two moles of carbon dioxide and three moles of water.
First, we need to calculate the number of moles of ethanol (C2H5OH) using its molar mass. The molar mass of ethanol is 46.07 g/mol.
Number of moles of ethanol = mass of ethanol / molar mass
Number of moles of ethanol = 340 g / 46.07 g/mol
Number of moles of ethanol ≈ 7.38 mol
Since the stoichiometric ratio between ethanol and oxygen is 1:3, the number of moles of oxygen required can be calculated as:
Number of moles of oxygen = 3 x number of moles of ethanol
Number of moles of oxygen = 3 x 7.38 mol
Number of moles of oxygen ≈ 22.14 mol
Now, we can use the ideal gas law to convert the number of moles of oxygen to liters.
PV = nRT
Assuming the reaction occurs at STP (standard temperature and pressure), we can use the following values:
Pressure (P) = 1 atm
Temperature (T) = 273 K
R = 0.0821 L·atm/(mol·K)
V = nRT / P
V = 22.14 mol x 0.0821 L·atm/(mol·K) x 273 K / 1 atm
V ≈ 502.42 L
Therefore, approximately 502.42 liters of oxygen are necessary for the combustion of 340 g of ethanol at STP.
To determine the number of liters of oxygen necessary for the combustion of ethanol, you need to follow a few steps:
Step 1: Write a balanced chemical equation for the combustion of ethanol. The balanced equation is:
C2H5OH + 3O2 → 2CO2 + 3H2O
This equation shows that one mole of ethanol reacts with three moles of oxygen to produce two moles of carbon dioxide and three moles of water.
Step 2: Calculate the number of moles of ethanol. To do this, you need to know the molar mass of ethanol, which is 46.07 g/mol. Divide the given mass of ethanol (340 g) by the molar mass:
Number of moles of ethanol = (mass of ethanol) / (molar mass of ethanol)
Number of moles of ethanol = 340 g / 46.07 g/mol ≈ 7.38 mol
Step 3: Determine the ratio of moles of ethanol to moles of oxygen. From the balanced equation, we can see that the ratio is 1:3. This means that every one mole of ethanol requires three moles of oxygen.
Step 4: Calculate the number of moles of oxygen needed. Multiply the number of moles of ethanol by the mole ratio:
Number of moles of oxygen = (number of moles of ethanol) × (mole ratio)
Number of moles of oxygen = 7.38 mol × 3 mol/mol ≈ 22.14 mol
Step 5: Convert the number of moles of oxygen to liters at STP (standard temperature and pressure). At STP, one mole of any gas occupies 22.4 liters. Use this conversion factor to find the volume:
Volume of oxygen = (number of moles of oxygen) × (22.4 L/mol)
Volume of oxygen = 22.14 mol × 22.4 L/mol ≈ 494.496 L
Therefore, approximately 494.5 liters of oxygen are necessary for the combustion of 340 g of ethanol at STP.
write the equation:
7O2 + 2C2H5OH>>4CO2 + 6H2O
check that.
so the moles of O2 is 3.5 x moles ethanol.
figure the moles of ethanol, then the moles of oxygen,then multiply by 22.4