132 grams of CH4 reacts with Oxygen gas to produce carbon dioxide and water vapor.
give the mass of the oxygen needed to react with all the methane and give the mass of the carbon dioxide and water vapor produced.
can somebody explain this to me? i'm lost. . .
You need to write the equation and balanc it. I assume you have it but didn't post it. Here is a step by step procedure for working stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Certainly! I'll break down the problem for you step by step.
To find the mass of oxygen needed to react with all the methane, we need to determine the stoichiometric ratio between CH4 and O2 in the reaction. The chemical equation for the reaction is:
CH4 + O2 -> CO2 + H2O
The balanced equation tells us that 1 molecule of methane (CH4) reacts with 2 molecules of oxygen (O2) to produce 1 molecule of carbon dioxide (CO2) and 2 molecules of water vapor (H2O).
To find the mass of oxygen needed, we first need to calculate the number of moles of methane. We can do this by using the formula:
Moles = Mass / Molar Mass
The molar mass of methane (CH4) can be calculated by summing up the atomic masses of its constituent elements. The atomic masses of carbon (C) and hydrogen (H) can be found on the periodic table.
C = 12.01 g/mol
H = 1.008 g/mol
Molar Mass of CH4 = (4 * H) + C = (4 * 1.008) + 12.01 = 16.043 g/mol
Now, we can calculate the number of moles of CH4 using the given mass:
Moles of CH4 = 132 g / 16.043 g/mol ≈ 8.22 mol
Since the stoichiometric ratio is 1 mole of CH4 to 2 moles of O2, we can conclude that 8.22 moles of CH4 will require 8.22 * 2 = 16.44 moles of O2.
Finally, to find the mass of oxygen needed, we multiply the number of moles by the molar mass of oxygen (O2). The molar mass of O2 is approximately 32.00 g/mol:
Mass of O2 = Moles of O2 * Molar Mass of O2 = 16.44 mol * 32.00 g/mol = 526.08 g
Therefore, 132 grams of CH4 will require around 526.08 grams of O2 to react.
To determine the mass of carbon dioxide (CO2) and water vapor (H2O) produced, we use the same stoichiometric ratio as before:
The moles of CH4 (8.22 mol) will produce an equal number of moles of CO2 and twice that number of moles of H2O.
So, Mass of CO2 produced = Moles of CH4 * Molar Mass of CO2
= 8.22 mol * (12.01 g/mol + 2 * 16.00 g/mol)
= 8.22 mol * 44.01 g/mol
= 362.07 g
And, Mass of H2O produced = 2 * Moles of CH4 * Molar Mass of H2O
= 2 * 8.22 mol * 18.02 g/mol
= 297.72 g
Hence, approximately 362.07 grams of CO2 and 297.72 grams of H2O will be produced.
I hope this explanation helps you understand the problem better!