What mass of carbon dioxide is produced from the complete combustion of 2.60×10−3 of methane?

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To find the mass of carbon dioxide produced from the complete combustion of methane, we first need to know the balanced chemical equation for the reaction.

The balanced equation for the complete combustion of methane is:

CH₄ + 2O₂ → CO₂ + 2H₂O

From this equation, we can see that for every 1 mole of methane (CH₄), we produce 1 mole of carbon dioxide (CO₂).

To find the mass of carbon dioxide produced, we need to convert the given amount of methane to moles, and then use the mole ratio from the balanced equation to find the moles of carbon dioxide. Finally, we can convert the moles of carbon dioxide to grams using the molar mass of carbon dioxide.

1. Convert the given amount of methane to moles:
To convert from mass to moles, we need to divide the mass by the molar mass of methane.

The molar mass of methane (CH₄) is:
(1 * Molar mass of Carbon) + (4 * Molar mass of Hydrogen)

The molar mass of carbon is approximately 12.01 g/mol, and the molar mass of hydrogen is approximately 1.01 g/mol.

So, the molar mass of methane is:
(1 * 12.01 g/mol) + (4 * 1.01 g/mol) = 16.04 g/mol

To convert the given mass of methane to moles, we divide the mass by the molar mass:
2.60×10^(-3) / 16.04 = 1.62×10^(-4) moles of methane

2. Use the mole ratio from the balanced equation to find the moles of carbon dioxide:
From the balanced equation, we know that 1 mole of methane produces 1 mole of carbon dioxide. Therefore, the number of moles of carbon dioxide is also 1.62×10^(-4) moles.

3. Convert the moles of carbon dioxide to grams:
To convert from moles to grams, we multiply the number of moles by the molar mass of carbon dioxide.

The molar mass of carbon dioxide (CO₂) is approximately 44.01 g/mol.

So, the mass of carbon dioxide produced is:
1.62×10^(-4) moles * 44.01 g/mol = 7.13×10^(-3) grams

Therefore, the mass of carbon dioxide produced from the complete combustion of 2.60×10^(-3) grams of methane is approximately 7.13×10^(-3) grams.