I have the compound C30H62, the candle

The question is:
When a candle is burned, 150.0 liters of O2 (g) is used up. How many grams of oxygen react?

22.4 liters/mol at stp

so how many mols?
150/22.4 approximately (it does not say if we are at stp)

grams of O2 = mols * 32

To determine the mass of oxygen that reacts when burning a candle with the given 150.0 liters of O2, we need to use the ideal gas law equation. The ideal gas law equation is:

PV = nRT

where:
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant
T = temperature

However, we are not given the pressure, temperature, or the number of moles directly. We can assume that the pressure and temperature are constant, as they are not mentioned explicitly. Let's rearrange the equation to solve for n:

n = PV / RT

Since we are given the volume of oxygen gas (V = 150.0 L), we need to convert it to moles of gas to use in the equation. To do this, we need to know the pressure and temperature of the gas, as well as the ideal gas constant (R).

Let's assume standard temperature and pressure conditions (STP) for this calculation. STP is defined as a temperature of 273.15 K and a pressure of 1 atmosphere (atm). The value of the ideal gas constant (R) is 0.0821 L·atm/(K·mol).

Now we can substitute the known values into the equation:

n = (150.0 L) * (1 atm) / (0.0821 L·atm/(K·mol)) * (273.15 K)

Simplifying this expression gives us the number of moles of oxygen gas (n). Since we know the molar mass of oxygen (O) is approximately 32 g/mol, we can multiply the number of moles by the molar mass to get the mass of oxygen that reacts:

Mass of oxygen = (n) * (molar mass of O)

Plug in the values and solve for the mass of oxygen.