What mass of oxygen is needed for the complete combustion of 2.50×10−3 of methane?

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.00249 g

To determine the mass of oxygen needed for the complete combustion of methane, we need to consider the balanced equation for the combustion of methane:

CH₄ + 2O₂ -> CO₂ + 2H₂O

From the equation, we can see that for every one mole of methane (CH₄) burned, we need 2 moles of oxygen (O₂).

First, let's calculate the number of moles of methane:

Molar mass of CH₄ = 12.01 g/mol (C) + 4(1.01 g/mol) = 16.05 g/mol

Number of moles of CH₄ = mass / molar mass
Number of moles of CH₄ = 2.50×10⁻³ g / 16.05 g/mol
Number of moles of CH₄ = 0.1558 mol

Next, based on the balanced equation, we know that the molar ratio of CH₄ to O₂ is 1:2.

Number of moles of O₂ = 2 * Number of moles of CH₄
Number of moles of O₂ = 2 * 0.1558 mol
Number of moles of O₂ = 0.3116 mol

Lastly, let's determine the mass of oxygen needed:

Molar mass of O₂ = 2(16.00 g/mol) = 32.00 g/mol

Mass of O₂ = number of moles of O₂ * molar mass of O₂
Mass of O₂ = 0.3116 mol * 32.00 g/mol
Mass of O₂ = 9.932 g

Therefore, the mass of oxygen needed for the complete combustion of 2.50×10⁻³ g of methane is 9.932 g.

To find the mass of oxygen needed for the complete combustion of methane, we first need to understand the balanced chemical equation for the combustion reaction of methane (CH4) with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).

The balanced chemical equation for the combustion of methane is:
CH4 + 2O2 -> CO2 + 2H2O

From this equation, we can see that one molecule of methane reacts with two molecules of oxygen to produce one molecule of carbon dioxide and two molecules of water.

To calculate the mass of oxygen needed, we follow these steps:
1. Determine the molar mass of methane (CH4).
2. Use the stoichiometric ratio from the balanced chemical equation to calculate the amount of oxygen needed.
3. Convert from moles of oxygen to grams of oxygen.

Let's calculate:

1. Molar mass of methane (CH4):
The molar mass of carbon is 12.01 g/mol, and the molar mass of hydrogen is 1.01 g/mol. Since methane (CH4) consists of one carbon atom and four hydrogen atoms, we can calculate the molar mass as:
(1 * 12.01 g/mol) + (4 * 1.01 g/mol) = 16.05 g/mol

2. Stoichiometric ratio:
From the balanced chemical equation, we can see that one molecule of methane requires two molecules of oxygen. Since we have 2.50×10−3 moles of methane, we can multiply it by the stoichiometric ratio to find the moles of oxygen required:
(2.50×10−3 mol CH4) * (2 mol O2 / 1 mol CH4) = 5.00×10−3 mol O2

3. Convert moles of oxygen to grams of oxygen:
To convert moles of oxygen to grams, we need to know the molar mass of oxygen. The molar mass of oxygen (O2) is approximately 32.00 g/mol. Now we can calculate the mass of oxygen:
(5.00×10−3 mol O2) * (32.00 g/mol) = 0.16 g

Therefore, the mass of oxygen needed for the complete combustion of 2.50×10−3 moles of methane is 0.16 grams.