# What volume of oxygen gas at 20.9 c and 750 torr is needed to burn 3.00 liters of propane, C3H8 at the same temprrature and pressure?

I know i use V1/n1=V2/n2, but how do i get n?

## write the reaction first.

C3H8 + 5O2 >> 3CO2 + 4H2O

so you need five times the volume of O2, looks like 15 liters.

Remember, in gases, at the same temp, pressure, contain the same number of moles per volume. Gay-Lussac's law of combining volumes applies

## To find the volume of oxygen gas needed, you can use the balanced equation for the combustion of propane:

C3H8 + 5O2 → 3CO2 + 4H2O

According to the balanced equation, for every one mole of propane (C3H8) burned, five moles of oxygen gas (O2) are consumed.

Given the initial volume, V1, and final volume, V2, you can use the formula V1/n1 = V2/n2 to calculate the number of moles involved.

In this case, the volume of propane is given as 3.00 liters. Since propane is in the gaseous state, it can be assumed that the volume is equal to the molar volume at the given temperature and pressure.

To calculate the number of moles of propane, you can use the ideal gas law equation:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

To convert the temperature from Celsius to Kelvin, you need to add 273.15 to the given value.

Once you calculate the number of moles of propane, you can use the stoichiometry of the balanced equation to determine the number of moles of oxygen gas required.

Using these steps, you can find the volume of oxygen gas needed by multiplying the number of moles of oxygen gas by the molar volume at the given temperature and pressure.

## To find the number of moles of oxygen gas needed to burn 3.00 liters of propane, C3H8, at the given temperature and pressure, you need to calculate the ratio of moles between propane and oxygen.

First, write out the balanced chemical equation for the combustion of propane:
C3H8 + 5O2 → 3CO2 + 4H2O

From the equation, you can see that 1 mole of propane reacts with 5 moles of oxygen gas. So the mole ratio between propane and oxygen gas is 1:5.

Since you have the initial volume of propane, you need to find the number of moles of propane by using the Ideal Gas Law equation:
PV = nRT

Here's the breakdown of the equation:
P = Pressure (in atmospheres)
V = Volume (in liters)
n = Moles
R = Ideal Gas Constant (0.0821 L.atm/mol.K)
T = Temperature (in Kelvin)

To convert the temperature from Celsius to Kelvin, add 273.15:
T (Kelvin) = T (Celsius) + 273.15

Now, calculate the number of moles of propane using the Ideal Gas Law equation:
n1 = (P1 * V1) / (R * T1)

Substitute the values given in the question:
P1 = 750 torr (convert to atm by dividing by 760)
V1 = 3.00 liters
R = 0.0821 L.atm/mol.K
T1 = 20.9°C (convert to Kelvin)

Calculate T1 (Kelvin):
T1 = 20.9 + 273.15

Plug the values into the equation to find n1:
n1 = ((750 / 760) * 3.00) / (0.0821 * (20.9 + 273.15))

Now that you have the number of moles of propane (n1), you can use the mole ratio to find the number of moles of oxygen gas (n2). Since the ratio is 1:5 (propane: oxygen gas), multiply n1 by 5 to find n2:
n2 = 5 * n1

Once you have the value for n2, you can proceed to find the volume of oxygen gas (V2) using the Ideal Gas Law equation:
V2 = (n2 * R * T2) / P2

Substitute the values given in the question:
n2 = moles of oxygen gas (calculated above)
R = 0.0821 L.atm/mol.K
T2 = 20.9°C (convert to Kelvin)
P2 = 750 torr (convert to atm by dividing by 760)

Calculate T2 (Kelvin):
T2 = 20.9 + 273.15

Plug the values into the equation to find V2:
V2 = (n2 * 0.0821 * (20.9 + 273.15)) / (750 / 760)

Finally, solve for V2 to find the volume of oxygen gas needed to burn 3.00 liters of propane at the given temperature and pressure.