What mass of oxygen is needed for the complete combustion of 2.10×10−3g of methane?
Express your answer with the appropriate units.
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To find the mass of oxygen needed for the complete combustion of methane, we need to write a balanced chemical equation for the combustion reaction of methane.
The combustion of methane (CH4) can be represented by the following equation:
CH4 + 2O2 → CO2 + 2H2O
From the balanced equation, we can see that one mole of methane reacts with two moles of oxygen to produce one mole of carbon dioxide and two moles of water.
First, we need to determine the number of moles of methane we have. Given the mass of methane as 2.10×10−3 g, we can use the molar mass of methane to convert this mass into moles.
The molar mass of methane (CH4) is the sum of the atomic masses of its constituent elements:
C: 12.01 g/mol
H: 1.01 g/mol (There are four hydrogen atoms in methane, so multiply by 4)
Molar mass of CH4 = 12.01 g/mol + (1.01 g/mol × 4) = 16.05 g/mol
Now we can convert the mass of methane to moles using the equation:
moles = mass / molar mass
moles of methane = 2.10×10−3 g / 16.05 g/mol ≈ 1.31×10−4 moles
Since two moles of oxygen are required for one mole of methane, we need to multiply the moles of methane by 2 to get the moles of oxygen needed.
moles of oxygen = 1.31×10−4 moles × 2 = 2.62×10−4 moles
Finally, we can convert the moles of oxygen to mass using the molar mass of oxygen, which is 32.00 g/mol:
mass of oxygen = moles of oxygen × molar mass
mass of oxygen = (2.62×10−4 moles) × (32.00 g/mol) ≈ 8.35×10−3 g
Therefore, approximately 8.35×10−3 grams of oxygen is needed for the complete combustion of 2.10×10−3 grams of methane.