What volume of O2 gas at STP is required to burn 32g of methane gas completely?
well, 1 mole occupies 22.4L
so, how many moles do you have?
what is the answer
To find the volume of O2 gas required to burn 32g of methane gas, we need to use the balanced chemical equation for the combustion of methane and the ideal gas law.
The balanced chemical equation for the combustion of methane (CH4) is:
CH4 + 2O2 -> CO2 + 2H2O
From the equation, we can see that 1 mole of methane reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and 2 moles of water.
First, let's calculate the number of moles of methane:
Molar mass of CH4 (methane) = 12.01 g/mol + 4*(1.008 g/mol) = 16.04 g/mol
Moles of CH4 = Mass of CH4 / Molar mass of CH4 = 32 g / 16.04 g/mol = 1.995 moles of CH4
Since 1 mole of methane reacts with 2 moles of oxygen, we need twice the number of moles of oxygen for complete combustion:
Moles of O2 = Moles of CH4 * 2 = 1.995 moles of CH4 * 2 = 3.99 moles of O2
Now, we can use the ideal gas law to find the volume of O2 gas at STP (standard temperature and pressure). STP is at a temperature of 273.15 K (0 degrees Celsius) and a pressure of 1 atm.
The ideal gas law is given by:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L*atm/(mol*K))
T = temperature (in Kelvin)
Since we know that 1 mole of any gas occupies 22.4 liters at STP, we can use this information to find the volume of O2 gas required.
Volume of O2 gas = Moles of O2 * 22.4 L/mol
Volume of O2 gas = 3.99 moles of O2 * 22.4 L/mol = 89.38 liters
Therefore, the volume of O2 gas required to burn 32g of methane gas completely at STP is approximately 89.38 liters.