When 16 grams of methane gas combine with 64 grams of oxygen, 44 grams of carbon dioxide form, plus water. What mass of water is produced?
You can work this problem as a stoichiometry problem OR you can recognize that the law of conservation of mass tells us that the mass on one side must equal to that of the other. First, make sure that all of the CH4 and all of the O2 are used.
To solve this problem using stoichiometry, we need to establish the balanced chemical equation for the reaction between methane (CH4) and oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).
The balanced chemical equation is:
CH4 + 2O2 -> CO2 + 2H2O
According to the equation, 1 mole of CH4 reacts with 2 moles of O2 to produce 1 mole of CO2 and 2 moles of H2O.
Next, let's convert the given mass of CH4 and O2 to moles using their respective molar masses.
The molar mass of CH4 (methane) is:
1 carbon (12.01 g/mol) + 4 hydrogen (4.01 g/mol) = 16.05 g/mol
The moles of CH4 can be calculated as:
moles of CH4 = mass of CH4 / molar mass of CH4
moles of CH4 = 16 g / 16.05 g/mol
moles of CH4 ≈ 0.996 moles (rounded to three decimal places)
The molar mass of O2 (oxygen) is:
2 oxygens (16.00 g/mol) = 32.00 g/mol
The moles of O2 can be calculated as:
moles of O2 = mass of O2 / molar mass of O2
moles of O2 = 64 g / 32.00 g/mol
moles of O2 = 2 moles
Since the ratio between CH4 and O2 in the balanced equation is 1:2, we can see that O2 is the limiting reactant because it can only react with 0.996 moles of CH4.
Now, let's calculate the moles of water produced. Since the balanced equation shows that 1 mole of CH4 produces 2 moles of H2O, we can use the mole ratio to find the moles of water.
moles of H2O = 2 * moles of CH4
moles of H2O = 2 * 0.996 moles
moles of H2O ≈ 1.992 moles (rounded to three decimal places)
Finally, let's convert the moles of water to mass using the molar mass of water.
The molar mass of H2O (water) is:
2 hydrogens (4.01 g/mol) + 1 oxygen (16.00 g/mol) = 18.02 g/mol
The mass of water can be calculated as:
mass of water = moles of H2O * molar mass of H2O
mass of water = 1.992 moles * 18.02 g/mol
mass of water ≈ 35.85 g (rounded to two decimal places)
Therefore, approximately 35.85 grams of water is produced.
To solve this problem using stoichiometry, you would start by writing a balanced chemical equation for the reaction:
CH4 + 2O2 -> CO2 + 2H2O
This equation shows that one molecule of methane (CH4) reacts with two molecules of oxygen (O2) to produce one molecule of carbon dioxide (CO2) and two molecules of water (H2O).
Next, you need to determine the number of moles of methane and oxygen used in the reaction. To do this, you can use the molar mass of each substance. The molar mass of methane (CH4) is 16 grams/mol, and the molar mass of oxygen (O2) is 32 grams/mol (16 grams/mol for each oxygen atom).
For methane:
Moles of CH4 = Mass of CH4 / Molar mass of CH4
Moles of CH4 = 16 grams / 16 grams/mol
Moles of CH4 = 1 mole
For oxygen:
Moles of O2 = Mass of O2 / Molar mass of O2
Moles of O2 = 64 grams / 32 grams/mol
Moles of O2 = 2 moles
Based on the balanced equation, you can see that 1 mole of CH4 reacts with 2 moles of O2 to produce 2 moles of H2O. Therefore, if all of the CH4 and O2 are used, you would expect to produce 2 moles of water.
Finally, to determine the mass of water produced, you can use the molar mass of water, which is 18 grams/mol. Multiply the number of moles of water by the molar mass of water:
Mass of water = Moles of water x Molar mass of water
Mass of water = 2 moles x 18 grams/mol
Mass of water = 36 grams
Therefore, a mass of 36 grams of water would be produced when 16 grams of methane gas combine with 64 grams of oxygen.