V(CH3OH)= 2 litres
ρ(CH3OH) = 0,80 g/ml
v(O2) - ?
(How much O2 is it needed to burn with 2 litres of CH3OH?)
To calculate the volume of oxygen (O2) needed to burn a certain amount of methanol (CH3OH), we need to determine the balanced chemical equation for the combustion of methanol.
The balanced chemical equation for the combustion of methanol is as follows:
2CH3OH + 3O2 → 2CO2 + 4H2O
From this equation, we can see that for every 2 moles of methanol (CH3OH) burned, we need 3 moles of oxygen (O2) to complete the reaction.
Now let's calculate the moles of methanol in 2 liters using the given density:
Density (ρ) of methanol (CH3OH) = 0.80 g/mL
To convert liters to milliliters, we multiply by 1000:
2 liters = 2 * 1000 = 2000 mL
Now, we need to calculate the mass of methanol using the density:
Mass of methanol (CH3OH) = Volume * Density = 2000 mL * 0.80 g/mL = 1600 g
Next, we need to convert the mass of methanol to moles using its molar mass:
Molar mass of methanol (CH3OH) = (12.01 g/mol * 1) + (1.008 g/mol * 3) + (16.00 g/mol * 1) = 32.04 g/mol
Number of moles of methanol (CH3OH) = Mass / Molar Mass = 1600 g / 32.04 g/mol ≈ 49.94 mol
Since the balanced chemical equation tells us that we need 2 moles of methanol (CH3OH) for every 3 moles of oxygen (O2), we can set up a proportion:
2 mol CH3OH / 3 mol O2 = 49.94 mol CH3OH / x mol O2
Cross-multiplying:
2 * x = 3 * 49.94
2x = 149.82
x ≈ 74.91
Therefore, approximately 74.91 moles of oxygen (O2) are needed to burn 2 liters of methanol (CH3OH).
To calculate the volume of oxygen (O2) in liters, we can use the ideal gas law:
PV = nRT
Assuming the temperature (T) and pressure (P) are constant, we can rearrange the equation to solve for volume (V):
V = nRT / P
Where:
n = number of moles of oxygen (O2)
R = ideal gas constant (0.0821 L·atm/mol·K)
P = pressure in atmospheres (atm)
Let's assume the pressure is 1 atmosphere (atm):
V = 74.91 mol * 0.0821 L·atm/mol·K * T / 1 atm
The temperature (T) can vary, so you would need to specify the temperature to find the exact volume of oxygen (O2) in liters needed to burn 2 liters of methanol (CH3OH).